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THE    ELEMENTS 


OF 


PLANE    TRIGONOMETRY 


BY 

WILLIAM   P.  DURFEE 

Professor  of  Mathematics  in  Hobart  College 


BOSTON,  U.S.A. 
GINN  &  COMPANY,  PUBLISHERS 

1901 


I 


Copyright,  1900 
By  WILLIAM  P.  DURFEE 

ALL  RIGHTS  RESERVED 


PEEFAOE 


In  preparing  this  book  the  author  had  two  ends  in  view : 

First,  to  give  the  student  an  elementary  knowledge  of 
the  science  of  Trigonometry,  together  with  an  introduction 
to  the  theory  of  functions  as  illustrated  by  the  trigono- 
metric ratios. 

Second,  to  give  him  practice  in  the  art  of  computation. 
Especial  stress  has  been  laid  on  this  side  of  the  subject  for 
the  reason  that  many  students  have  no  other  experience 
with  the  calculation  of  approximate  numbers.  The  value 
of  preliminary  estimates  of  results  and  the  necessity  of 
frequent  checking  are  constantly  insisted  on.  This  feature 
is  believed  to  be  novel. 

The  chapter  on  the  Right  Triangle  is  an  informal  intro- 
duction to  the  subject.  The  use  of  natural  functions  is 
advised  here  that  the  student  may  become  familiar  with 
them.  In  the  remainder  of  the  book  logarithms  are  used 
in  all  computations.  In  order  to  meet  what  seems  to  be 
the  demand  at  the  present  time,  the  author  has  worked  the 
illustrative  problems  with  five-place  logarithms.  Person- 
ally he  prefers  four-place,  as  they  are  sufficiently  accurate 
for  most  practical  purposes,  and  their  use  permits  the 
student  to  solve  more  problems  in  the  limited  time  at  his 


1 46543 


iv  PREFACE 

disposal.     By  dropping  the  fractional  part  of  the  minute 

and  the  fifth  figure  of  a  number,  where  they  occur  in  the 

data,  four-place  tables  may  be  used  without  trouble. 

The  author  is  under  great  obligation  to  Professor  E.  W. 

Davis,  of  the  University  of  Nebraska,  for  his  criticism  and 

suggestion.     Much  of  the  chapter  on  Computation  is  due 

to  him. 

W.  P.  D. 

Geneva,  August,  1900. 


CONTENTS 


CHAPTER    I  —  COMPUTATION 

SECTIONS 

1-2.  Approximate  Numbers   . 

3.  Logarithms     . 

4.  Use  of  Logarithmic  Tables 

5.  Interpolation 

6.  Accuracy  in  Computing 


CHAPTER   II  — THE    RIGHT   TRIANGLE 


7. 

Definition  of  the  Trigonometric  Functions 

8. 

Applications  of  the  Definitions 

9. 

Inverse  Functions  .... 

10. 

Functions  of  Complementary  Angles 

11-12. 

The  Fundamental  Relations  . 

13. 

Functions  of  45°,  30°,  and  60° 

14. 

Tables  of  Natural  Functions  . 

15. 

Use  of  Tables  of  Natural  Functions 

16-17. 

Solution  of  Right  Triangles    . 

18. 

Solution  of  Problems      .... 

CHAPTER   III  —  UNLIMITED   ANGLES 

19. 

Addition  and  Subtraction  of  Lines          . 

20. 

Angles 

21. 

Addition  and  Subtraction  of  Angles 

22. 

Measurement  of  Angles 

23. 

Quadrants 

24. 

Ordinate  and  Abscissa   . 

25. 

Definition  of  the  Functions    . 

26. 

Signs  of  the  Functions    . 

27. 

Fundamental  Relations 

28. 

Functions  of  0°,  90°,  180°,  270°      . 

29. 

Line  Representatives  of  the  Functions 

30. 

The  March  of  the  Functions  . 

81. 

Sine  Curve,  etc 

32. 

Periodicity     ..... 

VI 


CONTENTS 


CHAPTER   IV— REDUCTION  FORMULAS 

SECTIONS 

33.  Functions  of  Negative  Angles 

34.  Functions  of  90°  +  0,  etc.      . 

35.  Reduction  Formulas 

36.  Applications  ..... 


37-38. 

39. 

40. 
41-46. 

47. 

48. 

49. 

50. 

51. 

52. 


CHAPTER  V— THE  ADDITION 


THEOREM 


Projection       .... 
Projection  on  Coordinate  Axes 
Addition  Formula 
Sine  and  Cosine  of  <p  ±  d 
Tangent  and  Cotangent  of  <p  ± 
Functions  of  Double  Angles  . 
Functions  of  Half-angles 
Functions  of  Three  Angles     . 
Conversion  Formulas 
Trigonometric  Equations 


PAGE 

51 
52 
54 
55 


57 

59 

59 

60 

65 

67 

68- 

70 

71 

73 


CHAPTER  VI  — THE    TRIANGLE 

53.  Nomenclature,  etc.  .......       80 

54.  The  Law  of  Sines 81 

55.  The  Law  of  Tangents .82 

56.  The  Law  of  Cosines 82 

57.  Functions  of  Half -angles  in  Terms  of  Sides    .         .         .83 

58.  Circumscribed,  Inscribed,  and  Escribed  Circles      .         .       85 

59.  Area  of  the  Triangle 87 


CHAPTER  VII  — SOLUTION   OF   THE    TRIANGLE 

60.  Formulas  for  Solution  of  Triangles 

61.  Use  of  Logarithmic  Functions 

62.  A  Side  and  Two  Angles  .... 

63.  Two  Sides  and  the  Included  Angle 

64.  Two  Sides  and  the  Angle  Opposite  One  of  Them 

65.  Three  Sides , 


90 
93 
93 
94 
95 


APPENDIX 

Table  of  Important  Formulas 


103 


PLANE    TRIGONOMETRY 


CHAPTER    I 
ON   COMPUTATION 

1.  Suppose  I  measure  this  table  and  find  it  3  ft.  6  in.  long. 
Is  it  exactly  that  ?  not  a  shade  more  nor  less  ?  Obviously 
no  one  can  be  certain  of  this.  We  find  the  table  to  be  3  ft. 
6  in.  long  within  some  limit  of  accuracy  beyond  which  we 
do  not  care  or  are  not  able  to  go. 

Should  we,  by  the  refined  methods  known  to  science, 
attempt  to  get  the  length  to  within,  say,  a  millionth  of  an 
inch,  we  should  probably  find  that  two  different  measure- 
ments would  give  discordant  results,  while  neither  result 
would  agree  with  our  first  rude  measurement  of  3  ft. 
6  in. 

This  inaccuracy  holds  of  all  numbers  got  by  measure- 
ment ;  that  is,  with  the  great  bulk  of  numbers  with  which 
we  have  to  deal  in  practical  computation. 

Nor  can  we  altogether  avoid  this  approximation  when 
the  numbers  are  ideal.  For  example,  the  square  root  of  2 
is  1.4  to  the  nearest  tenth,  1.41  to  the  nearest  hundredth, 
1.4142  to  the  nearest  ten  thousandth ;  that  is  to  say,  the 
square  of  each  of  these  numbers  is  nearer  2  than  the  square 
of  any  other  with  the  same  number  of  places.  Similar 
remarks  apply  to  all  surds,  to  nearly  all  logarithms,  to  7r, 
and  to  the  various  trigonometric  ratios. 

1 


2  PLANE   TRIGONOMETRY 

It  is  plain  that  an  error  of  a  foot  in  a  mile  is  of  far  less 
relative  importance  than  an  error  of  an  inch  in  a  yard.  It 
is  indeed  a  wonderful  triumph  of  measurement  and  calcula- 
tion to  have  determined  the  sun's  distance  within  some  hun- 
dred thousand  miles,  the  whole  distance  being  92,900,000. 
This  being  the  accuracy  of  the  sun's  distance,  and,  further- 
more, the  earth's  orbit  being  only  approximately  circular,  it 
would  be  impossible  to  determine  the  length  of  the  orbit 
with  an  uncertainty  less  than  some  hundred  thousand  miles, 
even  though  we  knew  the  value  of  ir  to  a  million  places. 

2.  The  sum  of  a  number  of  approximate  numbers  cannot 
be  accurate  beyond  the  place  where  accuracy  ceases  in  any 
one  of  them. 

Suppose,  for  example,  two  men  had  measured  parts  of 
the  same  line,  one  finding  his  end  307.492  ft.  long  and  the 
other  his  end  602.43  ft.     The  length  of  the  whole  line  is 

307.492 

602.43 

909.92 

Since  the  second  man  did  not  measure  to  thousandths, 
the  result  cannot  be  accurate  beyond  hundredths. 

A  product  cannot  have  a  greater  degree  of  accuracy  than 
that  of  its  least  accurate  factor.  Suppose  we  wish  the  prod- 
uct of  the  approximate  numbers  23.57  and  612.3.  The 
approximate  number  23.57  may  have  any  value  from  23.565 
to  23.575,  while  612.3  lies  between  612.25  and  612.35. 
This  product  may  be  anything 

from  23.565  X  612.25  =  14427.67125 

to  23.575  X  612.35  =  14436.15125. 

The  mean  of  these  results  is  14432,  but  we  cannot  be 
sure  of  the  last  figure.     We  do  feel  sure,  however,  that  the 


23.57 

612.3 

14142. 

236.  * 

47.  ** 

7.  *  ** 

ON   COMPUTATION  6 

fourth  figure  is  nearer  3  than  any  other  digit.     The  product 
to  four  figures  is  14430. 

The  labor  of  obtaining  useless  figures  can  be  avoided  by 
simply  not  getting  those  partial  products  that  are  not  to 
be  retained  in  the  final  result. 

A  convenient  arrangement  is  as  below,  keeping  all  deci- 
mal points  in  line  and  multiplying  from  left  to  right.  The 
place  of  the  right-hand  figure  of  the  product 
of  the  multiplicand  by  any  multiplier  digit  is 
as  many  places  to  the  left  or  right  of  the  right- 
hand  figure  of  the  multiplicand  as  the  multi- 
plier digit  is  to  th&x  left  or  right  of  unit's  place. 
In  the  example  given  above,  2,  the  first  figure 
of  the  partial  product  arising  when  the  multi- 
plicand is  multiplied  by  6,  is  put  two  places  to  the  left  of  7, 
since  6  is  two  places  to  the  left  of  unit's  place.  The  stars 
indicate  places  of  figures  not  obtained  and  would  usually 
be  omitted.  We  carry  to  the  first  figure  retained  as  we 
would  carry  were  the  work  done  in  full ;  moreover,  if  the 
first  figure  omitted  is  5  or  more  than  5,  we  carry  1  to  the 
first  figure  retained.  This  is  illustrated  in  the  second  and 
the  fourth  partial  products  found  above. 

The  student  can  test  the  result  by  interchanging  multi- 
plier and  multiplicand.  If  the  multiplier  or  the  multipli- 
cand has  only  three-figure  accuracy,  it  will  readily  be  seen 
that  the  product  can  have  only  three-figure  accuracy. 

In  division  also  the  accuracy  of  the  quotient  cannot 
exceed  the  least  accurate  of  the  numbers  which  are  divi- 
dend and  divisor. 

The  arrangement  for  division  is  given  below. 

To  find  the  quotient  of  14432,  divided  by  612.3. 

Move  the  decimal  point  in  both  dividend  and  divisor  as 
many  places  to  the  right  as  are  necessary  to  make  the 


4  PLANE    TRIGONOMETRY 

23.57     divisor  an  integer.     The  left-hand  figure  of 

6123. 1 144320.         foe  quotient  is  placed  over  the  last  figure 

of  the  first  product,  and  the  decimal  point 

1837  of  the  quotient  is  in  line  with  the  decimal 

349  point  of  the  dividend.     After  the  last  sig- 

306  nificant   figure   of  the   dividend  has  been 

used,  the  partial  products  are  formed   by 

—  omitting   one,  then   two,  then   three    final 

figures  of  the  divisor,  remembering  to   carry  to  the  first 

figure  retained,  as  in  multiplication. 

3.  So  much  for  the  ordinary  arithmetical  processes.  The 
general  principle  underlying  it  all,  that  accuracy  of  results 
is  limited  by  accuracy  of  data,  continues  applicable  when 
tables  or  other  labor-saving  devices  are  used.  With  four- 
place  tables  seven-place  accuracy  is  not  to  be  looked  for, 
and  when  our  data  are  only  four-place  it  is  a  foolish  waste 
of  time  and  increases  the  liability  to  error  to  use  seven-place 
tables. 

LOGARITHMS    • 

The  labor  of  computation  is  very  greatly  abridged  by  the 
use  of  logarithms.  The  principal  facts  concerning  the  prac- 
tical use  of  logarithms  are  recapitulated  below : 

The  logarithm  of  a  number  is  the  power  to  which  10 
must  be  raised  to  produce  the  number.     Since 

10°  =  1,     101  =  10,     102  =  100,     103  =  1000,  etc., 
we  have  by  definition,  logarithm  1,  written 

log  1  =  0,  log  10  =  1,  log  100  =  2,  log  1000  =  3,  etc. 
If  m  and  n  are  any  two  numbers,  we  have  by  definition, 
m  =  10r,     or     log  m  =  x, 
n  =  10y,     or     log  n  =  y. 


ON   COMPUTATION  5 

Multiplying  these  two  equations, 

mn  =  \0x  +  y,     or     log  mn  =  x  -\-  y. 
.  * .  log  mn  —  log  m  +  log  n. 
Similarly, 

log  mnpq  •  •  •  =  log  ra  +  logm  -f-  log^>  +  log  q  H . 

I.  The  logarithm  of  the  product  of  several  numbers  is 
■equal  to  the  sum  of  the  logarithms  of  the  factors. 

Dividing  the  first  of  the  two  equations  above  by  the 
second, 

m       10'*'  m 

—  =  --—  =  lO*-2',     or     log  —  =  x  —  y  =  log  m  —  log  n. 

n ■       10y  n  J  .  & 

II.  The  logarithm  of. the  quotient  of  two  numbers  is  equal 
to  the  logarithm  of  the  dividend  minus  the  logarithm  of  the 
divisor. 

Raising  m  =  10*  to  the  -£th  power, 

mk  =  10**,     or     log  mk  —  kx  =  k  log  m. 

III.  The  logarithm  of  any  power  of  a  number  is  equal  to 
the  logarithm  of  the  number  multiplied  by  the  exponent  of 
the  power. 

Since  a  root  is  a  fractional  power,  the  logarithm  of  the 

kth.  root  of  a  number  is  -  times  the  logarithm  of  the  number. 

The  following  series  of  equations  illustrates  these  prin- 

€iples:  .JSP 

3  lab2       ,       fab2\h       1,       ah 


log  y  =  log\-^-  =  log  (  —  \    =  -  log  —  by  III 

—  i[log  ab2  —  log  c4]  by  II 

=  -J-  [log  <x  -j-  log  b2  —  log  c4]  by  I 

=  £[log  a  +  2  log  &  -  4  log  e]  by  III 


6  PLANE    TRIGONOMETRY 

4.  What  is  the  logarithm  of  22738  ?  Since  the  number 
lies  between  10000  and  100000,  its  logarithm  lies  between 
4  and  5.  By  calculation  it  has  been  found  to  be  4.35675. 
The  integral  part,  4,  is  its  characteristic ;  the  decimal  part, 
.35675,  is  the  mantissa.         « 


log  22738 

=  4.35675, 

or 

22738 

__   ^Q4.35675 

log  2273.8 

=  3.35675, 

or 

2273.8 

__   -^Q3.35675 

log  227.38 

=  2.35675, 

or 

227.38 

__   1Q2.356754 

log  22.738 

=  1.35675, 

or 

22.738 

_   ^Ql  .35675 

log  2.2738 

-  0.35675, 

or 

2.2738 

__  ^QO.35675 

log  .22738 

=  1.35675, 

or 

.22738 

_  ^Ql.35675 

log  .022738 

=  2.35675, 

or 

.022738 

—  ^02.^5675^ 

log  .002738 

=  3,35675, 

or 

.002738 

_  j[Q3i35675 

Inspection  of  this  algorithm  shows  us,  1st,  that  the  char- 
acteristic depends  solely  on  the  position  of  the  decimal 
point  and  can  always  be  determined  by  inspection;  2d, 
that  the  mantissa  is  independent  of  the  decimal  point  and 
depends  on  the  sequence  of  digits  constituting  the  number. 

The  student  can  easily  make  for  himself  a  set  of  rules 
for  determining  the  characteristic.  In  case  of  a  decimal, 
say  .000473,  he  can  determine  the  characteristic  of  473,  and 
then  move  the  point  six  places  to  the  left ;  by  so  doing  the 
characteristic  is  diminished  by  six  and  is  2  —  6  =  —  4, 
written  4.  The  minus  sign  is  written  above  the  character- 
istic because  it  is  the  characteristic  alone  that  is  negative, 
the  mantissa  being  always  positive'. 

The  mantissas  are  found  from  a  table.  Mantissas  are  all 
approximate  numbers,  and  tables  are  published  giving  man- 
tissas to  four,  five,  six,  and  seven  places.  The  kind  of  table 
to  use  depends  entirely  on  the  accuracy  of  the  numbers 
which  constitute  our  data.  Four-place  tables  are  accurate 
enough  for  ordinary  data  obtained  by  the  use  of  field  instru- 


ON   COMPUTATION  7 

ments,  five-place  tables  for  all  data  except  such  as  are 
obtained  by  the  use  of  the  most  delicate  instruments.  We 
shall  use  the  latter. 

USE    OF   THE    TABLE 

To  find  the  mantissa  of  2273  follow  down  the  left-hand 
column  of  your  table  of  the  logarithms  of  numbers,  passing 
from  page  to  page  until  you  reach  227  :  run  your  eye  across 
the  page  on  this  line  to  the  column  headed  3 ;  the  number 
so  reached,  35660,  is  the  mantissa  sought.  In  some  tables 
the  first  two  figures,  35,  are  printed  only  in  the  column 
headed  0. 

Verify : 

log  3748  =3.57380.  log  165     =2.21748. 

log  9741  =3.98860.  log  17       =1.23045. 

log  112.1  =  2.04961.  log  1624  =  3.21059. 

log  32.40  =  1.51055.  log  .0034  =  3.53148. 

5.  The  mantissas  of  five-figure  numbers  may  be  obtained 
from  the  table  by  interpolation.  The  mantissa  of  37423 
lies  between  the  mantissas  of  37420  and  37430.  We  assume 
that  it  lies  y\  of  the  way  from  the  first  mantissa  to  the 
second;  i.e.,  ^  of  the  way  from  57310  to  57322.  The 
difference  of  these  numbers  is  12  and  *fo  of  12  =  3.6  =  4. 
The  mantissa  of  37423  is  57310  +  4  =  57314. 

We  may  formulate  the  process  of  finding  the  mantissa 
of  a  five-figure  number  thus  :  Enter  the  table  with  the  first 
four  figures  ;  subtract  the  corresponding  mantissa  from  the 
next  larger  mantissa  to  find  the  tabular  difference  ;  multiply 
the  tabular  difference  by  the  fifth  figure,  considered  as  a 
decimal,  to  find  the  correction  ;  add  the  correction  to  the 
mantissa  first  found ;  the  result  is  the  mantissa  of  the  five- 
figure  number. 


8  PLANE   TRIGONOMETRY 

In  most  tables  the  multiplication  spoken  of  above  is  per- 
formed in  the  tables  of  proportional  parts  printed  on  the 
margin  of  the  page. 

Verify  the  following  logarithms  : 

log  127.34  =  2.10497.  log  8964.3  =  3.95252. 
log  34.876  -  1.54253.  log  90002  =  4.95425. 
log  .42748  =  T. 63092.         log  (.42748) 6  =  3.78552. 

The  last  problem  will  present  no  difficulty  if  we  remember 
that  the  mantissa  is  always  positive.  Division  of  the  log- 
arithm when  the  characteristic  is  negative  requires  care. 
Suppose  we  wish  to  divide  3.78552  by  6.  We  write  it 
6  -f  3.78552.  The  division  is  now  a  simple  matter.  If 
we  wish  to  divide  by  o,  we  write  it  5  +  2.78552.  We  make 
the  negative  characteristic  a  multiple  of  the  divisor. 

When  the  logarithm  is  given  and  we  wish  to  find  the 
number,  the  process  is  the  inverse  of  the  one  just  considered. 
We  may  formulate  it  thus  :  Find  in  the  table  the  mantissa, 
equal  to  or  next  less  than  the  given  mantissa.  The  corre- 
sponding number  will  be  the  first  four  figures  of  the  number 
sought.  Subtract  this  mantissa  from  the  given  mantissa 
to  find  the  correction.  Divide  the  correction  by  the  tabu- 
lar difference,  obtaining  a  one-figure  quotient.  Annex  this 
figure  to  the  four  already  found.  The  result  is  the  five- 
figure  number  corresponding  to  the  given  mantissa.  Place 
the  decimal  point  at  the  place  indicated  by  the  character- 
istic. The  result  is  the  number  corresponding  to  the  given 
logarithm. 

Verify  the  following : 

3.14216  =  log  1386.3.  2.37489  =  log  .023708. 

2.15362  =  log  142.44.  .96756  =  log    9.2803. 

1.87460  =  log    74.92. 


ON   COMPUTATION 


The   logarithm   of   —  is   called   the    cologarithm  of  my 

KYI 

written  either  colog  m  or  col  m.  In  computing,  it  often 
saves  labor  to  add  the  cologarithm  instead  of  subtracting 
the  logarithm. 

If  log  m  =  3.27463 

colog  m  =  log  1  —  log  m  —  0  —  3.27463 
=  6.72537  -  10. 

The  subtraction  is  readily  performed  from  left  to  right 
by  taking  each  digit,  except  the  last,  from  9.  The  —  10 
is  used  to  avoid  negative  characteristics.  Some  computers 
increase  all  negative  characteristics  by  10  and  take  account 
of  these  10's  in  the  final  result. 

6.  Accuracy  in  computing  can  be  attained  only  by  prac- 
tice and  by  constant  care.  When  the  computer  has  made 
his  interpolation,  he  should  glance  back  at  the  table  and 
see  that  his  result  lies  between  the  proper  tabular  numbers 
and  nearest  the  right  one.  This  takes  but  an  instant  and 
corrects  many  errors.  The  importance  of  carefully  plan- 
ning a  computation  before  entering  upon  it  can  hardly  be 
overestimated.  The  plan  should  be  written  out.  The  com- 
puter is  then  free  to  devote  his  whole  attention  to  the 
mechanical  details  of  the  work.  Paper  ruled  in  squares 
conduces  to  accuracy.  If  the  computation  be  confined  to 
one  column,  it  can  be  repeated  or  a  similar  one  inserted  in 
a  parallel  column  without  repeating  the  plan.  If  any  given 
number  occurs  repeatedly  in  a  computation,  it  may  be  written 
down  once  for  all  on  a  separate  piece  of  paper  and  held 
over  any  number  with  which  it  is  to  be  combined. 

The  computer  will  avoid  many  errors  if  he  accustoms 
himself  to  making  rough  estimates  of  results.     When  the 


/ 


10  PLANE   TRIGONOMETRY 

nature  of  the  subject  permits,  these  estimates  may  be 
obtained  by  graphic  methods. 

To  insure  accuracy  the  computer  must  continually  check 
his  work.  Every  operation,  every  step  in  every  operation 
must  be  tested  before  going  on.  If  two  numbers  are  added, 
subtract  one  of  them  from  the  sum.  If  two  numbers  are 
subtracted,  add  the  difference  to  the  smaller.  If  two  num- 
bers are  multiplied,  interchange  multiplier  and  multiplicand 
and  compare  products.  Test  every  step,  and  when  the  com- 
putation is  finished  check  the  final  result  if  the  nature  of 
the  problem  furnishes  a  test ;  if  not,  work  the  problem  by 
a  second  method  and  compare  results. 

The  computer  who  aims  at  rapidity  should  train  himself 
to  do  all  he  safely  can  mentally.  He  should  early  acquire 
the  habit  of  remembering  a  number  of  six  or  seven  figures 
long  enough  to  transcribe  it.  He  should  perform  his  inter- 
polations mentally.  He  should  add  and  subtract  two  num- 
bers from  left  to  right.  Other  devices  will  come  to  him 
with  practice. 

The  most  important  habit  to  be  acquired  is  that  of  being 
constantly  on  the  watch  for  errors  and  of  constantly  check- 
ing results.  The  computer  who  makes  no  mistakes  can 
hardly  be  said  to  exist.  Such  a  one  would  be  a  marvel. 
The  ordinary  man  who  forms  the  habit  of  not  letting  a  mis- 
take go  uncorrected  is  more  trustworthy  than  the  marvel 
who  does  not  verify  his  work. 


CHAPTEK   II 


THE   RIGHT   TRIANGLE 


7.    The  three  sides  x,  y,  r  of  the  right  triangle   ABC 
furnish  six  ratios : 

y    x    y    x     r    r 

—  9     —i     —9     —9     —  9     — • 

r    r    x    y    x    y 

If  a  second  right  triangle  A'B'C 
B   be  constructed  with  angle  A'  equal 
to  angle  A,  it  will  be  similar  to  the 
first  and  we  have  : 


x  :  y  :  r==  x  :  y  :  ?\ 
Therefore, 


Fig.  1. 


c' 

Jj/S 

/ 

V 

/^           x' 

A' 

B' 

Hi 


Fig.  2. 


y  x'  __  X  y  y 

v  r!       r  x'  x 

x'      x  r'       r  rf  r 

y*      y  x'      x  y]  y 


Each  ratio  of  one  triangle  is  equal  to  the  corresponding 
ratio  in  the  other. 

Let  us  construct  a  third  right  triangle  A"B"C",  making 
angle  A"  >  A  and  side  A"C"  =  AC. 
From  the  construction 


and 


r1!    __   ^ 

x' 

<  x, 

y' 

>m 

y"y 

"77  >  ~> 
r        r 

x' 

X 

<-> 

r 

x' 

X 

x"       X 

r     y 

x" 

r 
>-> 

X 

r' 

y" 
ii 

r 
<- 

y 

Fig.  3. 


12  PLANE   TRIGONOMETRY 

The  ratios  in  this  triangle  are  not  equal  to  the  corre- 
sponding ratios  in  the  first  triangle. 

The  foregoing  considerations  lead  to  the  conclusion  that 
these  ratios  depend  for  their  values  solely  on  the  angle  A  ; 
i.e.,  they  change  when  A  changes,  they  are  constant  when 
A  is  constant.  This  dependence  is  expressed  mathematic- 
ally by  saying  that  the  ratios  are  functions  of  the  angle  A. 
To  distinguish  them  from  other  functions  they  are  called 
Trigonometric  Functions. 

The    six    trigonometric    functions    of  A    are   named    as 
'  follows : 


-  =  sine  of  A , 
r 

written 

sin  A. 

-  =  cosine  of  A. 
r 

a 

COS^L. 

-  =  tangent  of  'A, 

a 

tan  A 

x 

-  =  cotangent  of  A 

y 

> 

cot  A. 

r 

-  =  secant  of  A, 

X 

a 

sec  A. 

r 

-^=  cosecant  of  .4, 

u 

esc  A. 

y 

8.'  The  foregoing  equations  define  the  trigonometric  func- 
tions. They  are  fundamental  and  should  be  carefully  mem- 
orized.    These  definitions  may  be  put  into  words  : 

?/      side  opposite 

sin  A  =  *  =  -r f^ 

r       hypotenuse 

x      side  adiacent 
cos  A  a=  -  r* 


r        hypotenuse 
y  _  side  opposite 
Fig.  4.  x      side  adjacent 


THE    RIGHT   TRIANGLE  13 


x      side  adiacent 

cot  A  =  -  =  — J- —• 

y      side  opposite 

_  r  _    hypotenuse 

x      side  adjacent 

r        hypotenuse 
esc  A  =  -  .—  — ^ 


t/      side  opposite 


EXERCISES 


Find  the  six  functions  of  each  of  the  acute  angles  in  the 
right  triangle  whose  sides  are : 


1.  5,  12,  13.  7.    a,  Vl  -  a2,  1. 

2.  3,  4,  5.  8.    a,  6,  Va2  +  62. 

3.  8,  15,  17.  v      9.    5,  5,  5  V2. 


4.  9,  12,  15.  10.    a,  V^ax~+ti\  a  +  x. 

5.  5,  8,  V89.  11.    a  +  b,  a-b,  V2(V  +  62). 

6.  2,  3,  Vl3.     ,  12.  m*—n2,  2mn,  m2  +  ^2. 

9.  Inverse  Functions.  Suppose  we  have  the  expression 
sin  .4  =  f;  how  may  we  describe  A  ?  A  is  the  angle  whose 
sine  is  f .     It  is  customary. to  express  this  by  writing 

A  =  sin-H. 

This  is  read  "A  is  the  angle  whose  sine  is  -J."     So,  too,  the 

expressions  cos-1 §,  tan-1|,  sec-1  f 

are  read  the  angle  whose  cosine  is 

§,  the  angle  whose  tangent  is  §, 

the     angle    whose    secant    is    f. 

These  functions  are  called  inverse 

functions.    They  are  distinguished   D  /<£         — ^ 

from  the  trigonometric  functions  FIG#  5# 

by  the  exponent  —  1. 


14  PLANE    TRIGONOMETRY 

Let  it  be  required  to  construct  sin-1!. 

Construction.  At  B  erect  BC  perpendicular  to  DB  and 
equal  to  3.  With  C  as  a  center,  and  with  a  radius  equal 
to  5,  describe  an  arc  cutting  BD  at  A.  Draw  CA  ;  then  is 
A  the  required  angle.     For  by  definition 

sin  A  =  | ,     or     A  =  sin-1 |. 

EXERCISES 
Construct  the  following  angles  : 

1.  sin-1 1,     sin-1  f,     sin_1f. 

2.  cos-1!,     cos-1 1,     cos""1!. 

3.  tan-1^,     tan-1 1,     tan-1l. 

4.  cot-1f,     cot-1 2,     cot-1 5. 

5.  sec-1 1,     sec-1  2,     sec-1 5. 

6.  esc-1 1,     esc-1 1,     esc-1 3. 
Show  by  constructing  a  figure  that : 

7.  sin-1f  =  cos-1  %  =  tan-1  §. 

Show  by  construction  that  the  following  angles  are  im- 
possible. 

8.  sin-1 1,     cos-1 2,     sec-1f,     csc-1^. 

9.  sm-1-7?    cos-1 7?    a>o. 

b  o 

10.    sec-1-?    esc-1-?    b<a. 
a  a 

10.  Functions  of  Complementary  Angles.  The  angles  A  and 
C  are  complementary.     By  definition 


V 

sin  A  =  -  =  cos  C. 
r 

cot  A 

x 

=  -  =  tan  C. 

y 

cos  A  =  -  =  sin  C. 
r 

sec  A 

r 

=  -  =  esc  C. 

X 

tan  A  =  -  =  cot  C. 

esc  A 

=  -  =  sec  C. 

THE   EIGHT   TRIANGLE 


15 


We  may  summarize  these  relations  by  saying  that  any 
function  of  an  angle  is  equal  to  the 
co-function  of  the  complementary 
angle.  In  this  statement  we  assume 
that  the  co-function  of  the  cosine  is 
the  sine,  etc.  The  cosine,  cotangent, 
cosecant  are  contractions  for  com  pie-  FlG-  6- 

ment's  sine,  complement's  tangent,  complement's  secant. 


EXERCISES 

1.  The  functions  of  30°  are  : 

sin  30°  =  ±,         cos  30°  =  i  V3,         tan  30°  -  £  V3. 
cot  30°  '=  V3,      sec  30°  =  §  V3,         esc  30°  =  2 ; 
write  the  functions  of  60°. 

2.  sin  40°  =  cos  50° ;  express  the  relations  between  the 
other  functions  of  these  angles. 

3.  The  angles  45°  -f  A  and  45°  —  A  are  complementary  ; 
express  the  functions  of  45°  -f-  A  in  terms  of  the  functions 
of  45°-  A. 

4.  A  and  90°  —  A  are  complementary ;  express  the  func- 
tions of  90°  —  A  in  functions  of  A. 

5.  45°  is  its  own  complement ;  show  that  sin  45°  =  cos  45°, 
tan  45°  =  cot  45°,'  sec  45°  =  esc  45°. 

11.    Fundamental  Relations  of  the  Trigonometric  Functions. 

The  six  functions  sine,  cosine,  etc., 
are  connected  by  a  number  of  equa- 
tions. The  more  important  of  these! 
are  derived  below.  The  first  five\ 
depend  immediately  on  the  defini- 
tions, the  other  three  on  a  well- 
known  property  of  the  right  tri- 
angle.    These  last  involve  the  squares  of  the  functions.    By 


Fig.  7. 


16  PLANE   TRIGONOMETRY 

universal  usage  powers  are  indicated  by  affixing  the  expo- 
nent to  the  functional  symbol.  E>g-,  (sin  A)2  is  written 
sin2  A,  (cos  A)z  is  written  cos3  A. 

y       v 
Since        -  X  -.  =  1  we  have  sin  A  esc  A  =  1  [1] 


cos  A  sec  A  =  1  [2] 

tan  A  cot  A  =  1  [3] 

sin  A 
cos  A 
cos  A 


V 

y 

X 

r 

X 

r 

X 

1 

y 

X 

X 

X 

y~ 

1 

y 

X 

y 

r 

r 

X 

X 

y  = 

X 

V 

r 

y 

sin  A 


tan  A  [4] 

=  cot  A.  [5] 


From  the  figure  y2  +  x2  =  r2. 

Dividing  this  equation  by  r2,  by  x2,  and  by  y2, 

S+S=1;    (r)+(7-)2=1-  •••SinM+cos^=l.[r,] 


=/ 


**?*■?/  i+©-w- 


These  eight  identities  constitute  the  fundamental  rela- 
tions of  the  trigonometric  functions.  They  are  very  im- 
portant and  should  be  committed  to  memory. 

By  means  of  these  relations,  when  we  know  one  function 
of  an  angle,  we  can  find  all  the  others. 

Suppose,  for  example,  sin  A  =  -J. 


[6]  cos  A  =  Vl  -  sin2 ,4  =  Vl  -  £  =  £  V& 

r  a  -.  sin  A  \  1  ,  .   r- 

[41  tan  ^4  = =  -~=z  =  --=  =  4  V3. 

■       cos.l      iV3       V3 


THE   RIGHT   TRIANGLE  17 


sin  A 


The  values  of  these  functions  may  also  be  found  by  con- 
structing sin-1£  and  finding  the  third  side  of  the1  triangle 
geometrically.  This  side  is  V3.  The  functions  can  now 
be  written  from  the  definitions  (§  8). 

EXERCISES 

Find  all  the  functions  of  the  following  angles,  using^each 
of  the  methods  illustrated  above  :  i 

1.  cot"1 2.         3.    cot"1!  5.    cos"1!!         7.    cos"1 2. 

2.  tan-1 3.         4.    sec-1  §.  6.    csc~^f.  8.    sin-1!. 

12.  We  can  express  any  function  of  A  in  terms  of  any 
other  function  of  A  by  making  use  of  formuras  [1]  to  [8]. 
As  an  illustration  let  us  express  each  of  the^functions  in 
terms  of  the  tangent. 


[3] 

tanvl  =  tan  A. 

cot  A  = -• 

tan  A 

[7] 

.  sec  A  =  Vl  +  tan2  A. 

1            i 

Vl  +  tanM 

[8] 
[1] 
[2] 

pc,0       A     -y/l         1       pn+2       |      "\/l         1                   A 

*         tan2  A 
1                 tan  A 

tan  A 

esc  A       Vl  +  tan2  A 

sec  A       Vl  +  tan2  A 

18 


PLANE   TRIGONOMETRY 


EXERCISES 

Express  each  of  the  functions  of  A  in  terms  of 

1.    sin  A.      2.    cos  A.      3.    cot  A.      4.    sec  .1.      5.    coo  A. 

6.  Tabulate  the  results. 

Prove  the  following  identities  by  means  of  formulas  [1] 
to  [8]. 

7.  sin  A  sec  A  =  tan  A. 

8.  (sin  A  +  cos  A)2  =  1  +  2  sin  A  cos  A. 

9.  (sec  A  4-  tan  A)  (sec  .4  —  tan  ^1)  =  1. 
1  —  sin  .4  _        cos  A 

cos  .4  1  +  sin  A 

11.  (l  +  tan^)24-(l  -tan.4)2  =  2  sec2,!. 

12.  (sin  A  4-  cos  ,4)2  4-  (sin  A  -  cos  ,4)2  =  2. 


13.    Functions  of  45°,  30°,  and  60°. 

Construct  angle  A  ==  45°,  lay  off  yli?  =  1,  and  complete 
the  right  triangle. 


Fig.  9. 

Angle  C  =  45°. 
From  the  definitions 

sin  45°  =  cos  45' 


Fig.  10. 
BC  =  1  and.4C  =  V2. 


V2 
tan  45°  =  cot  45°  =  1. 


sec  45°  =  esc  45° 


V2        r 


THE   RIGHT   TR.  NGLE  21 

Construct  the  equilateral  triangk  _  4452  =  26 
Bisect  angle  A  by  AB.    The  triangle  ii-_-jo9_-.o 
with  angle  BAC  =  30°,  angle  C  =  60°,~      '    ~~ 
side^=V3.  --4470- 

By  definition 

sin  30°  =  j.  sin  60°  =  -^  =  %V3. 

cos  30°  =  ^  =  £  V3.  cos  60°  =  j. 

tan  30°  =  -i=  =  i  V3.  tan  60°  =  — -  =  ^3. 

V3  1 

cot  30°  = =  V3.  cot  60°  =  — ^  =  4-  V3. 

1  V3 

sec  30°  =  -4=  =  f  V3.  sec  60°  =  f  =  2. 

V3      F  T 

esc  30°  =  *  =  2.  esc  60°  =  -^  .===  #  V& 

V3      3 

14.  Trigonometric  Tables.  In  the  preceding  paragraph  we 
have  found  the  functions  of  30°,  45°,  and  60°  by  simple 
geometrical  expedients.  The  functions  of  other  angles  are 
not  found  so  easily.  For  purposes  of  computation,  tables 
of  trigonometric  functions  are  used.  Such  tables  give  the 
values  of  the  sine,  cosine,  tangent,  and  cotangent  of  all 
angles  from  0°  to  90°  at  intervals  of  10'.  Examine  such 
a  table.  You  will  find  in  the  left-hand  column  the  angles ; 
their  sines,  cosines,  tangents,  and  cotangents  are  opposite  in 
appropriately  headed  columns.  The  column  at  the  extreme 
right  also  contains  angles.  Inspection  will  show  you  that 
these  angles  are  the  complements  of  the  corresponding 
angles  at  the  left.  We  learned  in  §  10  that  the  function  of 
any  angle  was  the  co-function  of  its  complementary  angle. 
The  sine   of   an  angle  at  the  right  is  the   cosine  of  the 


18  PLANE 

a   TRIGONOMETRY 


at  the  left.     You  will  find  this  indi- 

Express  each  of  iy  the  word  cosine  at  the  bottom  of  the 

1.    sin  A.      2.  Jie-     Similarly  for  the  other  functions.     If 

6.    Tabulate  /our  table  carefully,  you  will  find  that  the 

Pro^*  an  down  the  left  side  of  the  page  till  45°  is  reached ; 

they  then  run  up  the  right  side  of  the  page  till  90°  is 

reached.     If  the  angle  is  less  than  45°,  you  look  for  it  at 

the  left;  if  more  than  45°,  at  the  right.     If  the  angle  is 

at  the  left,  the  name  of  the  required  function  is  at  the  top 

of  the  page;  while  if  the  angle  is  at  the  right  the  name. of  the 

function  is  at  the  bottom  of  the  page. 

The  tables  do  not  contain  secants  and  cosecants.  These 
functions  are  reciprocals  of  cosine  and  sine,  and  can  readily 
be  found  by  taking  advantage  of  this  fact.  You  will  find 
by  experience  that  it  is  never  necessary  to  use  them  in 
computation. 

Take  your  table  and  run  down  the  column  of  sines.  They 
increase  with  the  angle.  So  do  the  tangents.  Examine  the 
cosines  and  cotangents.   They  decrease  as  the  angle  increases. 

15.  The  table  gives  the  functions  of  angles  which  are 
multiples  of  10f.  To  find  the  functions  of  other  angles 
we  interpolate,  as  explained  in  §  5.  Care  must  be  taken 
to  add  the  correction  in  finding  sines  and  tangents,  to  sub- 
tract it  in  finding  cosines  and  cotangents. 

Find  the  sine  of  27°  34'. 

sin  27°  30' -.4617. 
The  tabular  difference  =  .4643  -  .4617  ==  26. 

The  correction  =  .4  of  26  =  10.4  =  10. 

sin  27°  34'  =  .4617  +  10  =  .4627. 

Find  the  cosine  of  63°  27'. 

cos  63°  20'  =  .4488. 


THE   RIGHT   TRIANGLE  21 

The  tabular  difference  =  .4488  -  .4462  =  26. 

The  correction  -  .7  of  26  =  18.2  =  18. 

cos  63°  27'  =  .4488  -  18  =  .4470. 
Find  the  tangent  of  84°  28'. 

tan  84°  20'  =  10.078. 
The  tabular  difference  =  10.385  -  10.078  =  307. 

The  correction  =  .8  of  307  =  245.6  =  246. 

tan 84°  28'  =  10.078  +  246  =  10.324. 

The  work  which  is  here  done  out  in  full  should  be  per- 
formed mentally  as  far  as  possible.  In  case  your  table  has 
a  column  of  differences,  the  operation  of  finding  the  tabular 
difference  is  unnecessary ;  if  your  table  is  provided  with  a 
table  of  proportional  parts,  the  operation  of  finding  the 
correction  is  much  simplified. 

EXERCISES 
Verify  the  following : 

sin  0°  42'    =  .0122.  sin  58°  38'  =  .8539. 

cos  0°  42'    =  .9999.  cos  58°  38'  -  .5220. 

tan  0°  42'    =.0122.  tan  58°  38'  =  1.6405. 

co!;  9°  42'    ==  5.8505.  cot  58°  38'  =  .6096. 

sin  43°  01'  =  .6822.  cos  28°  13'  =  .8812. 

tan  38°  29'  =  .7949.  cot  81°  31'  =  .1492. 

To  find  sin-1. 4327. 

The  next  smaller  sine  is  .4305,  the  sine  of  25°  30'. 

The  difference  =  .4327  -  .4305  =  22. 

The  tabular  difference  =  .4331  -  .4305  ==  26. 

Correction  =  ||  =  8. 

.'.sin-1 .4327  =25°  38'. 


22  PLANE    TRIGONOMETRY 

Fin<i  cos"1  .8826. 

The  next  larger  cosine  is  .8829,  the  cosine  of  28°  00'. 

The  difference  =  .8829  -  .8826  =  3. 

The  tabular  difference  =  .8829  -  .8816  =  13. 

The  correction  *      =  T3^  =  2. 

.*.  cos-1  .8826  =  28°  02'. 

Here  the  next  larger  cosine  is  taken  because  the  cosine 
is  a  decreasing  function. 

EXERCISES 

Verify  the  following : 

tan-1  .4329  -  23°  24'.  tan"1  3.4268  =  73°  44'. 

cot"1  .3721  =  69°  35'.  cos"1     .4268  -  64°  44'. 

sin"1  .8523  =  58°  28'.  cot"1  1.4823  =  34°  00'. 

16.  The  Solution  of  the  Right  Triangle.  To  solve  a  right 
triangle  is  to  find  numerical  values  for  the  unknown  parts. 
This  is  possible  when  two  parts,  one  of  which  is  a  side, 
are  known.  Two  methods  of  solution  are  open  to  us,  —the 
Graphic  and  the  Trigonometric. 

Graphic  Solution.  It  is  desirable  to  solve  all  problems 
by  this  method  before  proceeding  to  the  more  accurate 
trigonometric  method.  It  gives  rough  approximations  and 
enables  the  student  to  detect  his  grosser  mistakes  in  the 
application  of  the  trigonometric  formulas.  The  solution 
consists  in  accurately  constructing  the  figure  from  the  data 
given.  The  required  or  unknown  parts  may  now  be  care- 
fully measured  by  scale  and  protractor.  The  use  of  paper 
ruled  in  squares  facilitates  this  work.  The  only  instru- 
ments needed  are  a  scale,  a  protractor,  a  straight-edge, 
and  a  pair  of  dividers.  Two-figure  accuracy  is  all  that 
should  be  aimed  at  in  this  method  of  solution. 


THE   EIGHT   TRIANGLE  23 

17.  Trigonometric  Solution.  In  using  this  method  we  com- 
pute the  values  of  the  unknown  parts.  The  first  four  defini- 
tions (p.  12)  furnish  formulas  sufficient  for  this  purpose. 

These  formulas  are : 

(a)  sin  A  =  J.   7  \  \  (C)  cos  A  =  y 

(b)  tan  A  =  ^.  (d)  cot  A  =  -• 

No  matter  what  two  parts  are  given,  one  of  these  four 
formulas  includes  them  both.  This  statement  assumes  that 
when  A  is  known  B  is  known,  since  the  two  angles  are 
complementary ;  and  it  further  assumes  that  when  A  is 
known  any  of  its  functions  are  known,  and  vice  versa, 
since  the  tables  enable  us  to  find  the  one  from  the  other. 

The  student  should  satisfy  himself  of  the  truth  of  this 
statement  by  selecting  all  possible  combinations  of  two 
parts  as  known  parts.  To  effect  the  solution  we  proceed 
as  follows : 

Select  a  formula  containing  the  two  known  parts,  and 
substitute  in  it  the  values  of  these  parts ;  the  resulting 
equation  will  give  a  third  part.  Of  the  three  parts  now 
known,  one  is  an  angle  and  two  are  sides.  To  find  the 
remaining  side,  select  a  formula  containing  it.  Where  pos- 
sible, a  formula  should  be  selected  which  does  not  contain 
the  computed  part.  Experience  will  show  that  this  is  pos- 
sible when  one  of  the  given  parts  is  an  angle. 

Checks.  These  computations,  like  all  others,  should  be 
checked.     A  convenient  formula  for  this  purpose  is 

r2  =  x2  -f-  y2, 
or  y2  =  r2  —  x2  =  (r  -f-  x)(r  —  x), 

or  x2  =  r2  —  y2  =  (r  +  y)(r  —  y). 

The  problems  that  follow  illustrate  the  process  of  solution. 


24 


PLANE   TRIGONOMETRY 


1°.    The  hypotenuse  of  a  right  triangle  is  36,  and  one  of 
its  angles  is  32°  14' ;  find  the  other  parts. 

Graphic  Solution.     Construct  the 

V      angle  A  =  32°  14',  or  as  nearly  so 

as  your  protractor  admits ;  lay  off 

K      AC  =  36;  drop  CB  _L  to  AB.     The 

triangle   ABC  is  the  required  tri- 

B      angle.     Measure  AB  and  BC. 


~^A 


Fig.  11. 
Trigonometric  Solution.      C=  90°  -  32°  14'  ==  57°  46'. 

Formula  (b)   cos  A   =  -  •     .  * .  x  =  r  cos  A . 
\  /  r 

[table]  x  =  36  (.8459)  =  30.45. 

Formula  (a)     sin  A  =  —  •     .'.  y  =  r  sin  ^4. 


[table] 
Check. 


y  =  36  (.5334)  =  19.20. 


(30.45)2  =  (55.20)  (16.80). 
927.2  =  927.4. 
This  shows  that  our  work  is  fairly  accurate. 

2°.    One  leg  of  a  right  triangle  is  27,  and  the  adjacent 
angle  is  67°  23' ;  find  the  other  parts. 

Graphic  Solution.  Construct  angle  A  = 
67°  23',  lay  off  AB  =  27,  erect  the  _L  BC. 
ABC  is  the  required  triangle.  Measure  AC 
and  BC. 

Trigonometric  Solution. 

C  =  90°  -  A  =  22°  37'. 
x 


Formula  (b)     cos  A  =  : 


x 


h 

r 

y 

L:^ 

Al    X=2T 

B 

21 
.3846 


cos  A 
=  70.21. 


Fig.  12. 


THE    RIGHT   TRIANGLE 


25 


Formula  (c)     tan  A  =  : 


y  ■ 


x  tan  A. 


Check. 


y  =  27  (2.4004)  =  64.81. 
y2  =  (r  +  #)  (7 —  a;). 
(64.81)2  -  (97.21)  (43.21). 
4200.2  =  4200.4. 


3°.    The  hypotenuse  of  a  right  triangle  is  48,  and  one  leg 
is  37 ;  find  the  other  parts. 

Graphic  Solution.  Construct  the 
right  angle  ABC,  lay  off  BA  =  37, 
from  A  as  center,  with  radius  48,  draw 
an  arc,  cutting  BC  in  C ;  draw  AC. 
ABC  is  the  required  triangle.  Meas- 
ure BC  and  the  angle  BAC. 

Trig ono metric  Solution. 


Fig.  13. 


Formula  (b) 


Formula  (a) 


Check. 


.7708. 

A  =39°  34'. 

C  =  90°  -  A  =  50°  26'. 


sin  A  ■= 


y 


~7a 


x=m 


.  y  =  48  sin  39°  34'. 

y  =  48  (.6370)  =  30.-58.  ' 

x2  =  (r  +  y)(r-  y). 
(37)2  =  (78.58)  (17.42). 
1369  =  1368.9. 


Fig.  14. 


4°.    The  two  legs  of  a  right  triangle  are 
j§   487  and  756;  find  the  other  parts, 
ss        Graphic   Solution.      In   the   right   angle 
ABC  lay  off  BA  =  487  and  BC  =  756;  draw 
_   AC.     A B C  is  the  required  triangle.     Meas- 
ure AC  and  the  angle  A. 


26 


PLANE   TRIGONOMETRY 


Trigonometric  Solution. 
Formula  (d)  cot  A  =  ± f  J  =  .6442. 

A  =  57°  13'. 

C  =  90°  -  A  =  32°  47'. 


Formula  (a)  sin  C  == 


487 


487 


487 


sinC      .5415 


Check. 


r 
r  =  899.4. 

y2  =  (r  +  x)  (r  —  x). 
(756)2  -  (1386.4)  (412.4). 
571536  =  571740. 
Four-figure  accuracy  is  all  that  we  expect,  and  this  we 
probably  have  in  r  but  not  in  (r  -f  x)  (r  —  x). 


EXERCISES 


Exercises  1-6  refer  to  Fig.  15 ;  7-16  refer  to  Fig.  16. 


1.  x  =  20,  r  =  30. 

2.  y  =  17,  r  =  60.  f  >/ 
j    3.    x  =  34,  y=  45. 


Fig.  15. 


7 


7.  a  =  20,     .4  =  30°. 

8.  ft  =  16,     ^  =  45°. 

9.  c  =  75,  .  J5  =  60°. 

10.  a  =  12,     ft  =  15. 

11.  a  =  407,  c   =  609. 


4.  A=  30°  24',  r  =  207. 

5.  c  =  38°  47',  r  =  103.4 

6.  A  =64°  23',  x=  20.32. 


Fig.  16. 

12.  ft  =  1.306,     c  =  2.501. 

13.  A  =  15°  17',  0  =  163. 

I  14.   ^  =  81°  17',  ft  =  .0143. 

15.  a  =  137.4,     ft  =  101.2. 

16.  5  =  65°  8',     c  =  3.145. 


/ 


THE   RIGHT   TRIANGLE  27 

^1§!  The  Solution  of  Problems.  The  problems  that  com- 
plete this  chapter  can  all  be  solved  by  right  triangles. 
While  different  problems  demand  different  methods  of 
solution,  the  following  general  method  of  procedure  will 
be  found  very  useful : 

1°.  Carefully  construct  a  diagram  to  some  convenient 
scale  and  find  the  graphic  solution  by  proper  measure- 
ments. 

2°.  Examine  the  diagram  for  a  right  triangle  with  two 
parts  given;  if  this  triangle  contains  the  required  part, 
solve  it ;  if  not,  consider  all  the  parts  of  this  triangle  as 
known,  and  find  another  right  triangle  with  two  parts 
known ;  if  this  second  triangle  contains  the  required  part, 
the  method  of  solution  is  obvious ;  if  it  does  not  contain 
the  required  part,  repeat  the  process  until  a  triangle  is 
found  that  does  contain  it.  It  may  be  necessary  to  draw 
auxiliary  lines.  When  you  have  found  the  several  steps 
that  lead  to  the  solution,  review  the  work  to  make  sure 
that  all  of  them  are  necessary. 

3°.  Proceed  to  the  computation,  being  careful  to  check 
each  step.  No  computation  should  be  made  until  the  whole 
process  of  solution  is  determined  upon  and  written  out. 

Definitions.  If  0  denote  an 
observer,  P  an  object  above  the 
horizon,  and  POH'  a  vertical 
plane  intersecting  the  horizon 
in  HH!,  the  angle  POH1  is  called 
the  Elevation  (or  Altitude)  of  P. 
If  the  object  be  below  the  hori-  FlG*  17# 

zontal  plane,  as  at  Q,  the  angle  QOH'  is  called  the  depression 
of  Q. 

The  bearing  of  an  object  is  its  direction  from  the  ob- 
server.    The  use  of  the  word  is  obvious  from  tfye  following 


28 


PLANE   TRIGONOMETRY 


illustrations.      If  0  be  the  observer,  the  bearing  of  P  is 
E  20°  N,  of  Q  is  N  25°  E,  of  R  is  W  30°  iV,  of  T  is  5  25°  W. 


Fig.  19. 

Bearing  is  also  often  given  in  terms  of  the  divisions  of 
a  mariner's  compass.  The  circle  is  divided  into  32  eqnal 
parts,  the  points  of  division  being  named  as  indicated* in 
the  figure. 

EXERCISES 

fly  The  center  pole  of  a  tent  is  20  ft.  high,  and  its  top  is 
stayed  by  ropes  40  ft.  long ;  what  is  the  inclination  of  the 
ropes  to  the  ground  ? 

\2J  A  man  standing  140  ft.  from  the  foot  of  a  tower  finds 
that  the  elevation  of  its  top  is  28°  25';  what  is  the  height 
of  the  tower  ?         ^. 

p  /m)  At  what  latitude  is  the  circumference 

oiuAparallel  of  latitude  equal  to  two-thirds 
'  of  the  circumference  of  the  equator  ? 
I         Suggestion.     Let  PEP'E'  be  a  section  of 
the  earth  through  its  axis,  PP'  its  axis, 
~p'  EE'  the  equator,  LL'  the   circle  of  lati- 

fig.  20.  tllde.     Then  will  EOL  be  the  latitude. 

4.  The  length  of  a  degree  of  longitude  at  the  equator 
is  69.16  mi.;  find  a  formula  for  the  length  of  a  degree  of 
longitude  at  latitude  A. 


THE    RIGHT   TRIANGLE  29 

/5j  A  ladder  40  ft.  long  reaches  a  window  33  ft.  from  the 
ground.  Being  turned  on  its  foot  to  the  opposite  side  of 
the  street,  it  reaches  a  window  21  ft.  from  the  ground; 
how  wide  is  the  street  ? 

(ey  From  a  window  the  top  of  a  house  on  the  opposite 
side  of  a  street  30  ft.  wide  has  an  elevation  of  60°,  while 
the  bottom  of  the  house  has  a  depression  of  30° ;  what  is 
the  height  of  the  house  ? 

/vN  A  pole  stands  on  top  of  a  knoll.  From  a  point  at  a  dis- 
tance of  200  ft.  from  the  foot  of  the  knoll,  the  elevations 
of  the  top  and  the  bottom  of  the  pole  are  60°  and  30°,  respec- 
tively ;  prove  that  the  pole  is  twice  as  high  as  the  knoll. 

8.  A  regular  hexagon  is  circumscribed  about  a  circle 
whose  radius  is  20  ft. ;  find  the  length  of  the  side  of  this 
hexagon. 

9.  The  radius  of  a  circle  is  1.  Find  the  side,  the  perim- 
eter, the  apothem,  and  the  area  of  a  regular  inscribed  poly- 
gon of  5  sides,  of  8  sides,  of  9  sides,  of  12  sides,  of  n  sides. 

10.  A  person  at  the  top  of  a  tower  100  ft.  high  observes 
two  objects  on  a  straight  road  running  by  its  foot.  The 
depression  of  the  nearer  is  45°  36',  of  the  more  remote  is 

^2£ ;  what  is  their  distance  apart  ?       x  <*^~ 

11.  If  the  edge  of  a  regular  tetrahedron  is  10  ft.,  find  the 
,  fy length  of  a  face  altitude ;  the  length  of  the  altitude,  and 

the  angle  between  two  faces. 

12.  The  roof  rises  from  the  adjacent  sides  of  a  square 
house  at  an  angle  of  30° ;  find  the  angle  which  the  corner 
of  the  roof  makes  with  the  horizon. 

13.  At  a  certain  port  the  seacoast  runs  X.  N.  E.,  and  a 
vessel  10  mi.  out  is  making  12  mi.  per  hour  S.  S.  W.  At 
2.30  p.m.  she  is  due  east ;  what  is  her  bearing  at  2  p.m.  ? 
at  3  p.m.  ?  At  2  p.m.  a  vessel  sailing  10  mi.  per  hour  is 
dispatched  to  intercept  her ;  what  course  must  the  latter 
vessel  take  ? 


30 


PLANE   TRIGONOMETRY 


Fig.  21. 


Suggestions.  Let  P  be  the  port  and  SS} 
the  course  of  the  vessel,  E  its  position  at 
2.30  p.m.  Let  PD  be  perpendicular  to  SS' ; 
find  when  she  will  be  at  D.  The  bearings 
at  2  p.m.  and  at  3  p.m.  will  be  easily  found. 
To  find  course  of  second  vessel  let  Q  be 
point  of  meeting  and  t  the  time  after  2 
p.m.,  when  they  meet.  PQ  and  DQ  can 
now  be  calculated  in  terms  of  t,  which 
can  be  easily  found. 

14.  A  smokestack  is  secured  by  wires  running  from 
points  35  ft.  from  its  base  to  within  3  ft.  of  its  top. 
These  wires  are  inclined  at  an  angle  of  40°  to  the  ground. 
What  is  the  height  of  the  smokestack  ?  the  length  of  the 
wires  ?  What  is  the  least  number  of  wires  necessary  to 
secure  the  stack  ?  If  they  are  symmetrically  placed,  how 
far  apart  are  their  ground  ends  ?  How  far  are  the  lines 
joining  their  ground  ends  from  the  foot  of  the  stack  ?  from 
the  top  of  the  stack  ?  What  angle  do  the  wires  make  with 
these  lines  ?  with  each  other  ?  What  angle  does  the  plane 
of  two  wires  make  with  the  ground  ?  What  angle  does 
the  perpendicular  from  the  foot  of  the  stack  on  this  plane 
make  with  the  ground  ?  what  is  its  length  ? 

15.  On  the  U.  S.  Coast  Survey  an 
observation  platform  50  ft.  high  was 
built.  The  platform  was  8  ft.  square. 
The  four  legs  spread  to  the  corners  of 
a  12-foot  square  at  the  base.  They 
were  braced  together  by  three  sets  of 
cross-pieces,  as  represented  in  the  illus- 
tration. If  the  cross-pieces  are  equidis- 
tant and  the  lowest  is  3  ft.  from  the 
ground,  find  the  length  of  each  piece 
required  for  the  construction. 


Fig.  22. 


THE   RIGHT  TRIANGLE  31 

y  16.  A  man  wishing  to  know  the  width  of  a  river  selects 
a  point,  A*  one  bank,  directly- 
opposite  a  tree,  TR,  on  the  other 
bank.  He  finds  its  elevation  to 
be  10°  30' ;  going  back  150  ft. 
to  B,  he  finds  its  elevation  to 
be  9°.  What  is  the  width  of  the 
river?     (Find  A  C,  perpendicu-  B^°A  Fiq  93 

lar  to  BR  ;  then  AR  and  A  T.) 

17.  Upon  the  top  of  a  shaft  125  ft.  high  stands  a  statue 
-   which  subtends  an  angle  of  3°  at  a  point  200  ft.  from  the 

shaft ;  how  tall  is  the  statue  ? 

18.  A  wheel  1  ft.  in  diameter  is  driven  by  a  belt  from 
a  wheel  4  ft.  in  diameter.  If  the  shafts  bearing  these 
wheels  are  parallel  and  10  ft.  apart,  how  long  will  the 
belt  be  (a)  if  crossed  ?   (/?)  if  not  crossed  ? 

19.  In  a  circle  whose  radius  is  15  ft.,  what  angle  will 
a  chord  of  20  ft.  subtend  at  the  center  ?  a  chord  of  25  ft.  ? 
of  10  ft.  ?  In  a  circle  of  radius  r,  what  angle  will  a  chord, 
a,  subtend  ? 

20.  In  a  circle  of  15  ft.  radius,  find  the  area  of  the  seg- 
ment cut  off  by  a  chord  of  18  ft. ;  the  area  of  the  segment 
included  between  this  chord  and  a  chord  of  25  ft. 

21.  The  base  of  a  quadrilateral  is  60  ft.,  the  adjacent 
sides  are  30  ft.  and  40  ft.,  the  corresponding  adjacent 
angles  are  110°  and  130°,  respectively ;  find  the  fourth 
side  and  the  other  two  angles. 

22.  The  elevation  of  a  balloon  due  north  from  A  is  60°  ; 
from  B,  1  mi.  west  of  A,  its  elevation  is  45° ;  what  is  the 
height  of  the  balloon  ? 

23.  One  of  the  equal  sides  of  an  isosceles  triangle  is 
47  ft.,  and  one  of  the  equal  angles  is  38°  24';  what  is 
the  base  of  the  triangle  ? 


; 


CHAPTER   III 
HE  TRIGONOMETRIC  FUNCTIONS  OF  UNLIMITED  ANGLES 

'  19.  A  directed  line  is  a  straight  line  generated  by  a  point 
moving  in  a  given  direction.  It  possesses  two  qualities  — 
length  and  direction. 

The  lines  AB  and  DC  are  of  equal  length  but  of  opposite 
direction  or  sign.  We  indicate  the  direction  of  a  line  by 
the  order  of  naming  its  extremities.     For  example : 


A B  AB  =  -BA, 

c_ <  D  or  AB  +  BA  =  0. 

Fig.  24. 

Parallel  directed  lines  may  be  added  by  placing  the 
initial  point  of  the  second  on  the  terminal  point  of  the 
first.  Their  sum  is  the  line  defined  by  the  initial  point 
of  the  first  and  the  terminal  point  of  the  second.  They 
may  be  subtracted  by  placing  their  terminal  points  together. 
The  remainder  is  the  line  defined  by  the  initial  points  of 
the  first  and  second. 


D E F, G 

Fig.  25. 
AB  +  BC  =  AC.  DE  +  EF  +  FG  =  DG. 

AC  +  CB  =  AB.  DG  -  FG  +  FE  =  DE. 

AC  -  BC  =  AB.  GF+  FE=  GE. 

AB  -  CB  =  AC.  GE-  FE  =  GF. 

32 


UNLIMITED   ANGLES  33 

The  difference  may  also  be  obtained  by  putting  the  initial 
points  together.  The  remainder  is  defined  by  the  terminal 
points  of  the  second  and  the  first. 

AC  -  AB  =  BC. 

AB  -  AC  =  CB. 

By  general  agreement  horizontal  lines  are  positive  when 
they  make  to  the  right,  negative  when  they  make  to  the  left. 
Vertical  lines  are  positive  when  they  make  upwards,  nega- 
tive when  they  make  downwards. 

Measurement.  A  directed  line  possesses  two  qualities  — 
length  and  direction.  Measurement  takes  account  of  both. 
Its  length  is  the  number  of  times  it  contains  the  unit  line, 
and  its  direction  is  indicated  by  its  sign. 

20.  Angles.  We  conceive  the  angle  L  VM  to  be  generated 
by  the  revolution  of  L  V  about  V  till  it  comes  into  coinci- 
dence with  VM.  This  revolution  may 
be  performed  in  two  ways :  1st,  as 
indicated  by  the  arrow  marked  a; 
2d,  as  indicated  by  the  arrow  marked 
J3.     In  a  the  motion  is  counter-clock-  IG'   6* 

wise  (opposite  to  the  motion  of  a  clock  hand),  in  f3  the 
motion  is  clockwise.  Mathematicians  have  agreed  to  call 
the  former  motion  positive.  The  angle  denoted  *by  a  is 
positive,  ft  is  negative. 

Nomenclature.  Capital  letters  denote  points  ;  small  let- 
ters, lines  ;  Greek  letters,  angles.  The  angle  above  may  be 
named,  indifferently,  LVM,  Im,  a. 

The  angle  ml  is  the  angle  described  when  m  turns  in 
a  positive  direction  to  coincidence  with  I.  The  angle 
described  by  m  when  it  turns  negatively  into  coincidence 
with  I  is  —  Im  =  —  a. 


34  PLANE    TRIGONOMETRY 

The  turning  line  is  the  initial  line  of  the  angle ;  the  other 
bounding  line  is  the  terminal  line.  In  naming  an  angle, 
the  initial  line  is  always  put  first. 

The  angle  Im  is  generated  by  the  revolution  of  I,  in  a 
positive  direction,  until  it  comes  into  coincidence  with  ra. 
If  I  continues  to  revolve,  it  will  again  come  into  coincidence 
with  m.  The  angle  it  has  described  is  still  called  Im.  It 
differs  from  the  former  Im  by  a  whole  revolution,  360°.  The 
two  angles  are  congruent.  If  I  continues  to  revolve,  it  will 
pass  m  repeatedly.  The  angles  described  when  it  passes 
m  are  all  denoted  by  Im.  They  differ  from  each  other  by 
some  multiple  of  360°.  They  are  congruent  angles.  While 
Im  denotes  any  one  of  these  congruent  angles,  the  smallest 
is  always  understood. 

The  student  may  get  a  clearer  conception  of  what  an  angle 
is  by  considering  the  motion  of  the  minute  hand  of  a  clock. 
In  one  hour  this  hand  describes  an  angle  of  360°;  in  an 
hour  and  a  half,  an  angle  of  540° ;  in  a  half  day,  an  angle 
of  4320°,  and  so  on.  At  12.15,  1.15,  2.15,  3.15,  etc.,  this 
hand  is  in  the  same  position.  Counting  from  12  o'clock,  it 
has  described  angles  of  90°,  450°,  810°,  and  1170°.  These 
angles  are  congruent.  It  is  to  be  remembered  that  in  the 
case  under  consideration  all  the  angles  are  negative. 

21.  Addition  and  Subtraction  of  Angles.  Angles  are  added 
and  subtracted  in  the  same  way  that  lines  are. 

The  sum  of  two  angles  is  found  by  placing  their  vertices 
together  and  bringing  the  initial  line  of  the  second  into 
coincidence  with  the  terminal  line  of  the  first,  preserving 
the  direction  of  both.  The  angle  determined  by  the  initial 
line  of  the  first  angle  and  the  terminal  line  of  the  second 
angle  is  their  sum. 

Two  angles  are  subtracted  by  bringing  together  their 
terminal  lines.     The  angle  determined  by  the  initial  lines 


UNLIMITED   ANGLES  35 

of  the  first  and  second  angles  is  their  difference.  The 
same  result  may  be  obtained  by  placing  their  initial  lines 
together.  Their  difference  is  defined  by  the  terminals  of 
the  second  and  the  first.  It  is  to  be  noted  that  the  differ- 
ence defined  above  is  the  difference  obtained  by  subtracting 
the  second  angle  from  the  first. 

L  VM  +  MVN  =  L  VN 

LVN+NVM  =  LVM.  n^^N 

L  VN  -  MVN  =  L  VM.  ^^n^^M 

L  VN  -  L  VM  =  MVN  ?&£— L 

Fig.  27. 
LVM-LVN  =NVM. 

22.  Measurement  of  Angles.  The  measure  of  an  angle  is 
the  number  of  times  it  contains  the  unit  angle,  and  this 
measure  will  be  positive  or  negative,  according  as  the  angle 
is  positive  or  negative. 

Definition.  A  Perigon  is  the  angle  generated  by  a 
single,  complete  revolution  of  a  line  about  a  point  in  a 
plane.     Two  unit  angles  are  in  common  use : 

The  degree,  which  is  3^  of  a  perigon.  This  unit  is  too 
familiar  to  require  further  comment. 

The  radian,  which  is  the  angle  whose  arc  is  equal  to  the 
radius. 

The  radian  is  a  definite  angle.  For  the  circumference  of 
any  circle  is  2  7r(7r  =  3.1416)  times  its  radius.  The  angle 
whose  arc  is  equal  in  length  to  the  radius  is  therefore  the 

angle  whose  intercepted  arc  is  - —  th  of  the  circumference. 

Z  7T 

Since  angles  are  proportional  to  their  arcs  the  radian  is 
- —  th  of  the  perigon. 

Z  IT 

Eadians  are  denoted  by  the  letter  r,  e.g.,  lr,  2r,  6r,  irr. 
Generally,  however,  this  symbol  is  omitted  unless  such 
omission  gives  rise  to  ambiguity. 


36  PLANE   TRIGONOMETRY 

Formulas  for  changing  from  degree  measure  to  radian 
measure,  and  vice  versa,  are  readily  obtained. 

2  7rr  =  360°. 

360°      180° 


-  ^_        2tt 

180° 
To  reduce  degrees  to  radians,  divide  by 

IT 

180° 
To  reduce  radians  to  degrees,  multiply  by 

IT 

When  the  angle  bears  a  simple  ratio  to  the  perigon,  its 
radian  measure  is  expressed  as  a  multiple  of  it. 

E.g.,  180° -7T,       90°  =  |,     270°-  |tt,     45°  =  J- 


EXERCISES 

1.  Express  the  following  angles  in  terms  of  it  radians  : 
30°,  45°,  60°,  75°,  90°,  105°,  120°,  135°,  150°,  165°,  210°, 
225°,  240°,  300°,  330°,  450°,  600°. 

2.  Express  the  following   angles   in  degrees  :    \  7r,  §  7r, 

t^  \**>  tV*j  A7^  to^  i77"- 

3.  Express  the  following  angles  in  radians  :  130°,  36°  4', 
147°  21',  200°,  340°  36',  38°  35'. 

4.  Express  the  following  angles  in  degrees  :  lr,  2r,  5'', 
lr.4,  3r.6,  5r.47,  8M,  10r,  1M. 

Degree  measure  is  used  in  all  practical  applications 
of  trigonometry,  while  radian  measure  is  used  in  analyt- 
ical work.  In  this  book  both  systems  are  used  indiscrimi- 
nately. 

23.  Quadrants.  It  is  customary  to  place  the  angle  in 
such  a  position  that  the  initial  line  is  horizontal  and  the 
vertex  of  the  angle  toward  the  left. 


II 

I 

0 

III 

IV 

UNLIMITED   ANGLES  37 

The  initial  line  and  a  line  through  the  vertex  perpen- 
dicular to  this  divide  the  perigon  into 
four  equal  parts  called  quadrants. 
These  quadrants  are  numbered  I,  II, 
III,  IV,  as  in  the  accompanying  figure. 
An  angle  is  said  to  belong  to,  or  to  b.e 
of,  the  quadrant  in  which  its  terminal 
line  lies.     Thus,  fig#  28. 

angles  >      0°  and  <    90°  are  of  the  1st  quadrant. 
"        >    90°   «     <  180°         "  lid        " 

"        >  180°    "      <  270°         "  Hid      " 

«        >270°    "     <360°         "  IVth     " 

Angles  greater  than  360°  may  be  said  to  belong  either 
to  the  quadrant  of  their  smallest  congruent  angle,  or  to  the 
quadrant  determined  by  counting  the  number  of  quadrants 
passed  over  in  the  generation  of  the  angle. 

E.g.,  the  angle  800°  is  congruent  to  80°,  since  800°  =  2x 
360°  +  80°. 

This  angle  belongs  to  either  the  1st  quadrant  or  to  the 
9th  since  800°  =  8x90°  +  80°. 

EXERCISES 

'  To  what  quadrant  do  each  of  the  following  angles  belong : 
50°,  150°,  200°,  300°,  400°,  500°,  600°,  700°,  1000°,  2000°, 
10000°,  100000°,  -  40°,  -  100°,  -  200°,  -  300°,  -  600°, 

3'  ~3~'  3^7r'  7^7r?  *7r'  <'7r?  5657r? 


24.  Ordinate  and  Abscissa.  The  position  of  any  point  in 
the  plane  is  uniquely  determined  as  soon  as  we  know  its 
distance  and  direction  from  each  of  the  two  perpendicular 
axes   XX'  and  YY'.      The  distance   from   XX'  (SP  in  the 


38 


PLANE    TRIGONOMETRY 


X 


s 


figure)  is  called  the  ordinate  of  the  point  P.  Its  distance 
from  YY1  (OS  in  the  figure)  is  the  abscissa  of  P.  Together 
they  are  the  coordinates  of  P.  The 
abscissa  is  usually  denoted  by  x,  and 
the  ordinate  by  y.  We  write  P  = 
(x,  y),  where  the  abscissa  is  always 
put  first.  These  coordinates  are 
directed  lines,  and  according  to  the 
convention  mentioned  in  §  19  (p.  33)y 
abscissas  which  make  to  the  right 
fig.  29.  are  positive,  to  the  left,  negative ; 

ordinates  which  make  upwards  are  positive,  downwards, 
negative.  The  initial  extremity  of  the  abscissa  is  on  the 
y-axis,  of  the  ordinate  on  the  rc-axis.  The  signs  of  the 
coordinates  in  the  several  quadrants  are  therefore : 


p2 

£, 

Y 
Rn 

* 

,   s3 

S4 

p3 

s2 

0 

$ 

T? 

PA 

4 

Y' 

X 

y 

i 

ii 

in 

IV 

+ 
+ 

+ 

— 

4- 

EXERCISES 

Draw  a  pair  of  axes  (preferably  on  coordinate  or  cross- 
section  paper)  and  fix  the  following  points  : 

3,  5  ;  -  3,  7  ;  3,  -  7;  -  6,  10  ;  4,  -  8  ;  -  3,  -  5  ;  -  1,  2  ; 
-  2,  -  6 ;  3,  0 ;  -  2,  0  ;  0,  4 ;  0,  -  5 ;  0,  0. 

Note.  Since  RP  —  OS  and  OR  =  £P,  the  coordinates  of  P  may 
be  taken  as  RP  and  OR  instead  of  OS  and  SP. 


25.    The  Trigonometric  Functions  of  any  Angle.     We  are 

now  in  a  position  to  define  the  trigonometric  functions  of 
any  angle.  These  definitions  are  more  general  than  those 
given  in  §  7  and  include  them. 


T  ^4-    7**i    /, 


UNLIMITED   ANGLES 


39 


Let  Im  (or  XOP)  be  any  angle.  Through  0,  its  vertex, 
draw  YY*  perpendicular  to  OX.  Take  XX]  and  YY'  as  axes 
of  coordinates.  Let  P  be  any 
point  on  m,  and  #,  ?/  its  coor- 
dinates. Let  OP  =  r,  and  let  us 
agree  that  r  shall  be  positive  . 
when  it  lies  on  m  and  negative 
when  it  lies  on  the  backward 
extension  of  m.  Denote  the 
angle  Im  or  XOP  by  <f>. 

The  functions  are  defined  as  follows : 

sine  of  cf>  =  sin  <j>  =  1/  /r. 

cosine  of  <£  =  cos  cf>  =  x/r. 
tangent  of  <£  =  tan<£  =  y  j  x. 
cotangent  of  <£  =  cot  cj>  =  x/y. 
secant  of  <£  =  sec<£  ~  v  /  x. 
cosecant  of  <f>    =  esc  <£  =  r/y. 

Note.  These  definitions  do  not  differ  from  those  in  §  7  except  in 
generality. 

These  are  all  the  possible  ratios  of  the  three  lines  x,  y, 
and  r. 

These  ratios  are  independent  of  the  position  of  P  on  m. 
For  if  P  be  taken  at  any  other  point,  as  P\  the  signs  of  x, 
y,  and  r  are  unchanged,  while  the  ratios  of  the  lengths  are 
the  same  in  both  cases,  since  the  triangles  OSP  and  OS'P' 
are  similar.  If  the  point  be  taken  at  Pn  on  the  backward 
extension  of  m,  the  signs  of  x,  y,  and  r  are  all  changed. 
The  triangles  OSP  and  OS"P"  are  similar.  The  ratios  of 
x,  y}  and  r  are  therefore  the  same  as  before  in  both  magni- 
tude and  sign. 


40 


PLANE    TRIGONOMETRY 


26.  These  ratios,  being  independent  of  the  position  of  P 
on  m}  are  functions  of  the  angle  <£.  Their  algebraic  signs 
depend  upon  the  quadrant  to  which  <j>  belongs. 

Draw  an  angle  of  each  quadrant  and  verify  the  following 
table,  taking  r  positive. 


X 

y 

sin 

cos 

tan 

cot 

sec 

CSC 

I 

+ 

+ 

+ 

+ 

+ 

+ 

+ 

+ 

II 

— 

+ 

+ 

— 

— 

— 

— 

+ 

III 

— 

— 

— 

— 

+ 

+ 

— 

— 

IV 

+ 

— 

— 

■f 

— 

— 

+ 

— 

In  quadrant  I  all  functions  are  positive. 

"  II  "  negative  except  sin,  esc. 

"         III  "  "  tan,  cot. 

"  IV  "  "  cos,  sec. 

EXERCISES 

1.  Write  down  the  signs  of  the  several  functions  of  the 
following  angles : 

40°,  100°,   160°,   200°,  250°,    300°,   340°,   -40°,   -80% 
-  130°,  -  190°,  -  240°,  -  300°. 

2.  In  Fig.  31  the  lines  CC  and  BB!  are  drawn  equally 
Y  inclined  to  XX',  forming  the  angles 

^yC    <£u  <ta  <t>3,  <£4-     If  we  take 

OP1  =  OI\  =  OPs  =  OP4, 

the  coordinates  of  the  points  Plr 
P2,  P3,  and  P4  will  be  equal  in  mag- 
nitude but  not  in  sign.  We  shall 
have 

sin  <£x  =  sin  <f>2  =  —  sin  <£3  =  —  sin  <£4. 
Find    the    corresponding    relations    between    the   other 
functions. 


UNLIMITED   ANGLES  41 

3.  Show  by  Fig.  31  that  there  are  always  two  angles  less 
than  a  perigon  which  have  the  same  sine :  1st,  when  the 
sine  is  positive ;  2d,  when  it  is  negative.  Show  the  same 
thing  for  each  of  the  other  functions. 

4.  Construct  the  following  angles  :   (See  §  9.) 
sin-1§;         sec-1 3;         cos-1  — i;     tan-1  — 2; 
cot-1  — 1;     sec-1  — 2;     sin-1  —  i ;     csc-1J; 
tan-1 2;         cos-1§;  sin-1§;  esc-1 3. 

Remember  that  in  each  case  there  are  two  solutions. 

5.  Prove  the  following  equations  by  means  of  a  diagram  : 
sin  60°    =  sin  120°;  tan  225°  =  tan  45° ; 

cos  30°    =  -  cos  150° ;  cos  45°    ==  sin  135° ; 

cos  120°  =  -  sin  30° ;  tan  150°  =  -  tan  30° ; 

sec  40°    =  -  sec  140° ;  cot  130°  =  -  cot  50° ; 

sin  210°  =  -  sin  30° ;  tan  135°  =  -  tan  45°. 

6.  What  angle  has  the  same  sine  as  35°,  130°,  190°, 
350°,  47°,  -  40°,  -  140°,  -  230°,  -  340°,  cj>  ?  What  angle 
has  the  same  cosine  as  each  of  the  preceding  angles  ?  the 
same  tangent  ?  the  same  cotangent  ?  the  same  secant  ?  the 
same  cosecant  ? 

7.  Draw  a  diagram  and  find  the  functions  of  120°. 
(See  §13.) 

8.  Find  the  functions  of  each  of  the  following  angles : 
135°,  150°,  210°,  225°,  240°,  300°,  315°,  330°.      -^ 


27.    Fundamental  Relation  of  the  Trigonometric  Functions. 

The  relations  [1]  to  [8],  which  were  proved  in  §  11  for 
acute  angles,  can  be  readily  shown  to  hold  for  all  angles. 
The  proof  is  left  for  the  student.  For  convenience  of 
reference  they  are  repeated  here. 


42 


PLANE   TRIGONOMETRY 


sin  <£  •  esc  <j>  =  1. 

[i] 

cos  <£  •  sec  cf>  =  1. 

[2] 

tan<£-  cot  4>  =  1. 

[3] 

,        ,       sin  <£ 
tan  <b  = • 

COS  cf> 

[4] 

cos  <£ 

COt  d>  —     .          - 

■       sm  cj> 

[5] 

sm2cf>  +  cos2<£  =  1. 

.       [6] 

1  +  tan2<£  ==  sec2<£. 

m 

1  +  cot2  cj>  =  CSC2  cj>. 

[8] 

The  following  scheme  may  assist  in  remembering  the 
first  three  of  these  formulas : 

sin  </> 


cos  <f> 

tan  <f> — , 
cot  <f> ' 

sec  <£ 

esc  <£ 


=  1. 


EXERCISES 

By  means  of  the  relations  [1]  to  [8]  verify  the  following 
equations : 

1.    sin  <£  =  tan  <f>  cos  <£.        4.    cos  <£  =Vl  —  sin2<£. 
tan  0  sin  <j> 


2.    sin  <j>  = 


sec  <£ 


5.    tan  cj> 


vT^  sin2<£ 


3.    sin  <£  =  Vl  —  cos2$.       6.    tan  <£  ="Vsec2<£  —  1. 


7.    cot  <f>  = 


Vl  -  sin2<fr 
sin  <£ 


cos  <£ 


V 1  —  cos2  </>      tan  cj> 

1 
Vsec2<£  —  1 


=  Vcsc2<£  —  1. 


UNLIMITED   ANGLES 


43 


8. 

9. 

10. 

11. 
12. 

13. 

14. 
15. 
16. 
17. 
18. 
19 
20. 


Express  each  of  the  functions  in  terms  of  the  sine. 
cos2<£  —  sin2<£  =  1  —  2  sin2<£  =  2  cos2<£  —  1.    \ 
sec2<£  +  csc2<£  ==  sec2<£  csc2</>.  ^, 

sin  <£       _  1  qp  cos  <£ 
1  ±  cos<£  sin  (j> 

cos  <£       _  1  ±  sin  </> 
1  qz  sin  <f>  cos  4> 

sec  <£  ±  1  _       tan  <£ 

tan  <£  sec  cf>  zp  1 

tan  <£  +  cot  <£  =  sec  <£  esc  <£. 

sin  $  =  i',  find  all  the  other  functions  analytically, 
cos  <£=—$;      "  "  «  " 

tan<£  =  §;  «  «  "  " 

sec  <£  =  f ;  "  "  "  " 

sin  <£  +  cos  <£  ==  1.2 ;   find  sin  cf>.  *-^ 
tan2<£  —  sin2<£  =  tan2</>  sin2<£. 


2|.    The  functions  of  0°,   90°,  180°, 
270°,  360°. 

Let  P  be  a  point  on  the  terminal  line  h 


sin  0°  =  -  =  0, 


Z 


of  <£  at  a  distance  r  from  the  origin.        p3 

When  <j>  =  0,  P  coincides  with  the 
point  P1?  and  its  coordinates  are  x  =  r,  Pj- 

y  =  0.  Fig.  32. 


cos  0°  =  -  =  1, 


tan  0°  =  -  =  0, 
r 


cot  0°  =  -  =  oo, 


sec  0°  =  -  =  1, 
r 


esc  0°  =  -  =  oo. 


44  PLANE   TRIGONOMETRY 

When  <£  =  90°,  P  coincides  with  P2,  and  its  coordinates 

ace  x  =  0,  y  =■  r. 

sin  90°  =  -  =  1,     cos  90°  =  -  =  0,     tan  90°  =  £  =  oo, 
r  r  0 

cot  90°  =  -  =  0,     sec  90°  ==J  =  oo,    esc  90°  =  -  =  1. 
r  .      0  r 

When  <£  =  180°,  P  coincides  with  P3,  and  its  coordinates 
are  x  =  —  r,  y  =  0. 

sin  180°  =  -- 0,  cos  180°  =  —  =  -  1, 

r  r 

tan  180°  ==  —  =  0,  cot  180°  =  ^—  =  oo, 

sec  180°  =  —  -  -  1,  esc  180°  =  £  =  oo. 

—  r  0 

When  <\>  =  270°,  P  coincides  with  P4,  and  its  coordinates 
are  x  =  0,  y  =  —  r. 

sin  270°  =  —  =-  1,  cos  270°  -  -  =  0, 

r  r 

tan  270°  =  ^  -  oo,  •  cot  270°  =  —  =  0, 

0  —  r 

sec  270°  =  £  =  qo,  esc  270°  =  —  =  -  1. 

0  —  r 

When  <£  =  360°,  P  coincides  with  Plf  and  the  functions 
of  360°  are  identical  with  those  of  0°. 

It  is  customary  to  prefix  a  double  sign  to  the  zero  and 
infinity  values  of  the  functions,  the  upper  sign  being  that 
of  the  function  in  the  preceding  quadrant,  the  lower  that 
of  the  function  in  the  following  quadrant. 

The  results  obtained  are  tabulated  in  the  first  table  on 
the  opposite  page. 


UNLIMITED   ANGLES 


(3 


OS 
cos  L OP  =  —  =  length  of  OS,  ' .'  OP  =  unit  of  length. 

SP       LT 

cot  LOP  =  -^-  =  i7—  =         "       NH,'.'ON  =  " 

SP       ON 

OP       OT 
sec  LOP  =  — =  —  =       «        OT,\'OL=  « 

OS       OL 

csc  LOP  = = =        "         OH'.'ON  =  « 

SP       ON 

If  we  agree  that  secants  and  cosecants  shall  be  positive 
when  measured  on  the  terminal  line  and  negative  when 
measured  on  the  backward  extension  of  this  line,  it  will 
be  found  on  examination  that  these  lines  represent  the 
functions  in  sign  as  well  as  in  magnitude.  For  example, 
L  T,  the  tangent,  is  positive  in  quadrants  I  and  III,  negative 
in  quadrants  II  and  IV. 

30.  The  March  of  the  Functions.  We  will  now  study  the 
variation  or  march  of  each  of  the  several  functions  as  the 
angle  increases  from  0°  to  360°.  As  the  angle  increases 
the  point  P  travels  in  the  positive  direction  around  the 
'  circumference  of  the  circle.  As  P  passes  through  the  1st, 
2d,  3d,  and  4th  quadrants : 

The  sine,  SP,  increases  from  0  to  1,  decreases  to  0, 
decreases  to  —  1,  increases  to  0. 

The  cosine,  OS,  decreases  from  1  to  0,  decreases  from 
0  to  —  1,  increases  to  0,  increases  to  1. 

The  tangent,  LT,  increases  from  0  to  go,  changes  sign 
and  increases  from  -co  to  0,  increases  to  oo,  changes  sign 
and  increases  from  —  oo   to  0. 


48 


PLANE    TRIGONOMETRY 


The  cotangent",  NH,  decreases  from  oo  to  0,  decreases  to 
—  co,  changes  sign  and  decreases  from  oo  to  0,  decreases 
to  —  oo    and  changes  sign. 

The  secant,  OT,  increases  from  1  to  oo,  changes  sign  and 
increases  from  —  oo  to  —  1,  decreases  from  —  1  to  —  oo, 
changes  sign  and  decreases  from  oo   to  1. 

The  cosecant,  OH,  decreases  from  oo  to  1,  increases  from 
1  to  oo,  changes  sign  and  increases  from  —  oo  to  —  1, 
increases  from  —  1  to  —  oo   and  changes  sign. 

These  results  are  tabulated   below  : 


sin 

0° 

1st  Quad. 

90° 

2tl  Quad. 

180° 

3d  Quad. 

270° 

4th  Quad. 

0° 

=f° 

inc. 

1 

dec. 

±0 

dec. 

-1 

inc. 

*o 

cos 

1 

dec. 

±0 

dec. 

-1 

inc. 

*o 

inc. 

1 

tan 

T° 

inc. 

±oo 

inc. 

=F° 

inc. 

±oo 

inc. 

T° 

cot 

If  00 

dec. 

±0 

dec. 

zpoo 

dec. 

±Q 

dec. 

qpoo 

sec 

1 

inc. 

±oc 

inc. 

-1 

dec. 

^00 

dec. 

1 

esc 

=F°° 

dec. 

1 

inc. 

±oo 

inc. 

-1 

dec 

ipoo 

31.  Graphic  Representation  of  the  Functions.  The  nature 
of  the  variations  which  we  have  just  been  studying  may  be 
exhibited  by  the  following  constructions. 

Divide  the  circumference  of  the  unit  circle  into  any  num- 
ber of  equal  parts.  In  the  figure  the  points  of  division  are 
marked  0, 1,  2,  3  ...  12.  Lay  off  the  same  number  of  equal 
parts  on  a  horizontal  line,  and  number  the  points  of  division 
in  the  same  way.  Make  the  divisions  of  the  line  approxi- 
mately equal  to  the  divisions  of  the  circumference. 

At  the  points  0,  1,  2,  3  on  the  line  erect  perpendiculars 
equal  in  sign  and  length  to  the  sine  (SP)  of  the  correspond- 
ing point  on  the  circle.  Join  the  ends  of  these  perpendicu- 
lars by  a  continuous  line.     The  resulting  curve  is  the  curve 


UNLIMITED   ANGLES 


49 


of  sines.  As  P  moves  along  the  circle,  SP  changes  continu- 
ously, i.e. ,  it  changes  from  one  value  to  another  by  passing 
through  all  intermediate  values.     If  now  we  conceive  S1  as 


3 

i< 

f 

t 

► 

* 

\  / 

9<r 

\180° 

270° 

300  y 

0 

L         2 

i 

>         o\     7 

8 

9 

10 

11 

/13 

S 

V 

^ 

M 


Fig.  34. 

moving  along  LM,  keeping  pace  with  P.  while  S'P'  is  equal 
to  S'P,  the  point  P'  will  trace  the  curve  of  sines.  Our  con- 
struction is  an  attempt  to  realize  this  conception. 

32.  If  the  angle  increases  beyond  360°,  i.e.,  if  P  makes 
a  second  revolution,  the  values  of  SP  would  repeat  them- 
selves in  the  same  order.  If  we  plot  these  values,  we  shall 
have  the  curve  between  L  and  N  repeated  beyond  2V,  and 
this  curve  will  be  repeated  as  many  times  as  P  makes 
revolutions.     The  sine  curve  will  take  this  form.     (Fig.  35.) 

The  student  should  construct  the  curve  of  cosines,  the 
curve  of  tangents,  and  the  curve  of  secants  in  a  similar 
manner.  To  find  the  tangents  and  secants,  the  construc- 
tion of  the  preceding  section  should  be  used.  The  sum  or 
the  difference  of  two  functions  may  be  plotted.     To  plot 


50 


PLANE    TRIGONOMETRY 


sin  cj>  -f  cos  <£,  erect  at  0,  1,  2,  3,  etc.,  on  MN  (Fig.  33), 
perpendiculars  equal  to  SP  +  OS  at  the  several  points  0, 1, 
2,  3  on  the  fundamental  circle  in  that  figure.  The  result- 
ing curve  will  represent  sin  <£  +  cos  <£. 


( 

) 

T 

2 

7T 

37T 

4 

IT 

5 

7T 

6 

7T 

0 

/        ^\ 

X 

+  \ 

( 

)°          li 

10° 

360° 

540° 

720° 

900° 

1080 

Fig.  35. 

Periodic  Functions.  Functions  which  repeat  themselves 
as  the  variable  or  argument  increases  are  called  periodic 
functions.  The  period  is  the  amount  of  change  in  the. 
variable  which  produces  the  repetition  in  the  values  of 
the  function.  The  sine,  as  is  evident  from  Fig.  35,  is  a 
periodic  function  Avith  a  period  of  360°,  or  2  it.  The  tan- 
gent has  7r  for  its  period. 


EXERCISES 

Plot  the  following  functions  and  determine  their  periods  : 

1.  sin  <£  —  cos  <£.  4.    sin(90o-h  <£). 

2.  tan<£  —  sin  <f>.  5.    sin(—  <£). 

3.  sec  <f>  —  tan<£.  6.    cos(—  <£). 

7.    cos  <£  and  sec  <£  on  the  same  axes. 


CHAPTEK   IV 
REDUCTION   FORMULAS 


33.  Negative  Angles.  The  object  of  this  chapter  is  to 
obtain  a  set  of  formulas  which  will  enable  us  to  express 
any  function  of  an  angle  greater  than 
90°  as  a  function  of  an  angle  less 
than  90°. 

Let  AOC  be  a  negative  angle  and 
AOC'  an  equal  positive  angle.  Lay  off 
OP  =  OP'  and  draw  PP'.  SP'  =  —  SP. 
Let  the   coordinates  of  P  be  x,  y,  of 


p'  =  x\  y. 

Now  x  =  x',  y  =  - 

-y'. 

sin  (—  d>)  =  -  = 

•  sin  <£. 

X 

COS(—  d>)  =  -  = 
J       r 

x'  _ 
r 

cos<£. 

tan  (—(b)  =  -  = 

v         J        X 

£ 

x' 

-  tan  cj>. 

X 

COt  ( —  d>)  =  -  = 

y 

x' 

-  cot  <£. 

sec(—  d>)  =  -  = 

v                      X 

r 
x' 

sec  </>. 

V 

csc(—  d>)=  -  = 

y 

r 

-y'~ 

-  CSC  <f>. 

[9] 


A  little  reflection  will  show  that  this  proof  is  independ- 
ent of  the  magnitude  of  <£  and  is  therefore  general.     Its 

51 


52 


PLANE   TRIGONOMETRY 


results  may  be  summed  up  by  saying  that  in  passing  from 
—  <£  to  <£  the  functions  do  not  change  name  but  do  change 
sign  except  the  cosine  and  secant. 


34.    Let  AOB  =  cf>  and  A OC  =  90°  +  ,<£.     Lay  off  OP'  = 
OP.     The    two    triangles    OPS   and    OP'S'    are    congruent 

(equal).      Let    coordinates    of   P 
be  x,  y\    of  P'f  x'y'.     Neglecting 
algebraic  signs,  we  have 
x  =  y',  x'  =  y. 

No  matter  what  the  magnitude 
of  cj>,  it  is  obvious  that  we  shall 
always  have  this  relation  between 
the  coordinates  of  P  and  P'. 

Fig.  37.  «-./..,.  -,  ^ 

By  definition  we  have,  neglect- 
ing algebraic  signs, 


^ 

Y 

\P 

A/ 

S 

<x'v 

0 

y 

I 

p' 

Yl 

sin  (90°  +  $)  =  ~  =  ~  =  cos  <£. 
cos  (90°  +  <£)  =  -  =  ^  =  sin  <£. 


tan  (90°  +  <£)  =  ^  =  -,  =  cot  <£. 
x        y 


cot(90°  +  4,)  =  ^: 


tan  <f>. 


v       v 

sec  (90°  +  <£>)=  —  =  -  =  esc  <f>. 
v  J      x'      y 

esc  (90°  +  <£)  =  -,  =  -  =  sec  <f>. 
y       x 


By  studying  this  table  (p.  53)  of  the  signs  of  the  functions 
in  the  several  quadrants,  it  appears  that  sin  (90°  -f-  <£)  and 
cos  <£  always  have  the  same  algebraic  sign.    For  if  (90°  +  <£) 


REDUCTION   FORMULAS  53 


falls  in  quadrant  IV,  <£  falls  in  quadrant  III,  n          i 

and  sin  (90°  4-  <j>)   and   cos  <£  are  both   neg-  sin  + 

ative.    If  (90°  +  <£)  falls  in  III,  <f>  falls  in  II,  cos  - 

and  both  are  negative,  etc.      We  conclude :  tan ' 


SID   + 

cos  + 
tan  + 
cot  + 

the  cosine  of  an  angle  in  the  preceding  quad-    cgc   ■    !  csc   , 
rant  have  the  same  algebraic  sign. 


The  sine  of  an  angle  in  any  quadrant  and 


sin  — 
cos  + 
tan  — 
cot  — 
sec  + 

CSC  — 


sin  — 
Cos  (90  +  <£)  and  sin  <j>  have  different  signs.    cos  ._ 

For  the  sign  of  the  cosine  in  any  quadrant  tan  + 

in  the  table  is  different  from  the  sign  of  the  cot  + 

sine  in  the  preceding  quadrant.  sec  ~~ 

By  employing  the  same  method  of  reason-  ~" 
ing  we  can  show  that 

tan  (90°  4-  </>)  and  cot  <£  have  different  signs 

cot  (90°  4-  <t>)     "  ■  tan  <£     "            "  " 

sec  (90°  4-  <£)    "    csc  <j>     "           "  " 

csc  (90°  4-  </>)     "    sec  <t>     "     the  same  " 

The  preceding  formulas  (p.  52),  written  with  the  proper 
signs,  are:  sin(90o  +  (^)=      cos  ^ 

^  cos  (90°  4-  4>)  =  —  sin  ^. 

tan  (90°  4-  <£)  =  -  cot  <£.  [10] 

cot  (90°  +<£)  =  -  tan  <£. 

sec  (90°  4-  <£)  =  -  csc  </>. 

csc  (90°  4-  </>)  =      sec  <£. 
Since  180°  4-  <£  =  90°  4-  90°  4-  <k 

sin  (180°  4-  4>)  =      cos  (90°  4-  <£)  =  -  sin  <£. 

cos  (180°  4-  <f>)  =  -  sin  (90°  4-  <t>)  =  -  cos  <£. 

tan  (180°  4-  <f>)  =  -  cot  (90°  +  <A)  =      tan<£.        [11] 

cot  (180°  4-  <£)  =  -  tan  (90°  4-  <£)  =      cot  </>. 

sec  (180°  +  <£)  =  —  csc  (90°  4-  <£)  =  -  sec  <£. 

csc  (180°  +  <#>)=      sec  (90°  4-  </>)  -  -  csc  <£. 


54  PLANE    TRIGONOMETRY 

Functions  of  270°  +  <£  are  found  by  putting  270°  +  £ 
=  180°  +  90°  +  <£•  The  results  are  tabulated  below.  The 
student  is  advised  to  verify  these  results  by  drawing  dia- 


grams. 

A 

90° -0 

90° +  0 

180°  -  0 

180°  +  0 

270°  -  0 

270°  +  0 

360°  -  0 

sin 

COS  <£ 

COS  </> 

sin  <f> 

—  sin  <j> 

—  cos  <£ 

—  cos_<£ 

—  sin  <f> 

cos 

sin  <£ 

—  sin  cj> 

—  cos  <f> 

—  COS  <f> 

—  sin  <£ 

sin  </> 

cos  <£ 

tan 

cot  <j> 

—  cot  <£ 

—  tan</> 

tan  <j> 

cot  <£ 

—  cot  cj> 

—  tan<£ 

cot 

tan<£ 

—  tan  </> 

—  cot  <f> 

cot  <£ 

tan<£ 

—  tan<£ 

—  cot  <j) 

sec 

CSC  <£ 

/sec  <j> 

—  sec  cf> 

—  sec  <f> 

—  CSC  <j) 

CSC  <£ 

sec  <j> 

CSC 

sec  <£ 

CSC  <j> 

—  esc  <£  —  sec  <j> 

—  sec  <f> 

—  CSC  <f> 

This  table  includes,  beside  the  cases  we  have  already 
discussed,  the  functions  of  90°  -  <j>,  180°  -  <£,  270°  -  <£, 
and  360°  —  <£.      These  are  reduced  as  follows : 

sin  (90°-<£)=sin  [90°+(-<^)]=    cos(-^)=     cos<£,by[9] 
sin(180o-^)=sin[180°+(-<#))]=-sin(-^)=     sin  <£,     " 
sin(270o-^)=sin[270°+(-^)]  =  --cos(-^)  =  -cos  <£,    « 
sin(360°-<£)=sin  (-^)=-sin  <£. 

The  other  functions  of  these  angles  are  derived  in  a 
similar  manner. 

35.  If  we  inspect  the  table  carefully,  we  find  that  it  can 
be  summed  up  in  the  two  rules  that  follow : 

1°.  If  90°  or  270°  is  involved,  the  function  changes  name 
(from  sine  to  cosine,  from  tangent  to  cotangent,  from  secant 
to  cosecant,  and  vice  versa),  while  if  180°  or  360°  is  involved 
the  function  does  not  change  name. 

The  second  rule  has  to  do  with  the  algebraic  sign. 

When  we  write 

cos  (90°  +  <£)    =  —  sin  <£, 
tan  (180°  -£)==—  tan  <£, 


REDUCTION   FORMULAS  55 

iDoth  terms  must  have  the  same  sign.  If  <£  is  less  than  90°, 
sin  <£  is  positive  and  cos  (90°  -4-  <£)  is  negative.  The  equality- 
is  secured  by  putting  the  minus  sign  before  sin  <f>.  Since 
these  formulas  are  general,  the  signs  are  the  same,  no  matter 
what  the  value  of  cf>.     Our  rule  is  then : 

2°.    Assume  that  <£  is  less  than  90°  and  make  the  signs 
of  both  terms  alike.     ^ 


36.    Applications.     Any  angle  greater  than  90°  can   be 
expressed  in  two  of  the  following  forms : 
90°+<fc  180°-<£,  180°+ <£,  270°  -  cf>,  270°+4>,  360° -<fc 
where  <f>  is  less  than  90°. 

E.g.,  200°  =  180°  +  20°,  or  270°  -  70°. 

300°  =  270°  +  30°,  or  360°  -  60°. 
135°  =    90°  +  45°,  or  180°  -  45°. 

The  functions  of  200°  are : 

sin  200°  =  sin  (180°  +  20°)  =  -  sin  20°. 
cos  2.00°  =  cos  (180°  +  20°)  =  -  cos  20°. 
tan  200°  =  tan  (180°  +  20°)  =  tan  20°. 
cot  200°  =  cot  (180°  +  20°)  =  cot  20°. ' 
sec  200°  =  sec  (180°  +  20°)  =  -  sec  20°. 
esc  200°  =  esc  (180°  +  20°)  =  -  esc  20°. 

They  may  also  be  written : 

sin  200°  =  sin  (270°  -  70°)  =  -  cos  70°. 
cos  200°  =  cos  (270°  -  70°)  =  -  sin  70°. 
tan  200°  =  tan  (270°  -  70°)  =  cot  70°. 
cot  200°  =  cot  (270°  -  70°)  =  tan  70°. 
sec  200°  -  sec  (270°  -  70°)  =  -  esc  70°. 
esc  200°  -  esc  (270°  -  70°)  =  -  sec  70°. 


56  PLANE   TRIGONOMETRY 

EXERCISES 

1.  Express  the  following  functions  as  functions  of  angles 
less  than  90°:  tan  130°,  sin  160°,  cos  100°,  cot  215°,  sec260°, 
esc  280°,  sin  310°,  cos  310°. 

•     2.    Express  each  of  the  preceding  functions  as  the  func- 
tion of  an  angle  less  than  45°. 

3.  Express  each  of  the  following  functions  as  the  func- 
tion of  an  angle  less  than  45°     =  —    • 

.      2  7T  3  7T  5lT  47T  ,    5  77  11  7T 

sm— ?  tan——?  cos— — >  sec  -=->  cot  — ^- >  esc     .    • 
3  4  6  3  3  6 

4.  By  using  formulas  [9]  express  the  following  functions 
as  functions  of  positive  angles  less  than  90° : 

sin  (-160°),    cos  (-30°),    tan  (-300°),    sec  (-140°), 
cot  (-240°),    esc  (-100°),    sin  (-300°). 

5.  The  angle  —  <£  is  obviously  congruent  to  360°  —  <j>, 
and  their  functions  are  identical.  Reduce  the  functions  in 
problem  4  by  making  use  of  this  identity.. 

6.  Express  the  following  functions  as  functions  of  angles 
less  than  45° : 

cos  117°  17',    sin  143°  21'  16",    tan  317°  29'  31", 
cot  90°  46'  12",    sec  (-  135°  14'  11"),    cos  (-  71°  23'). 


CHAPTER  V 
THE   ADDITION  FORMULA 

37.  Projection.  The  projection  of  &  point  on  a  line  is  the 
foot  of  the  perpendicular  from  the  point  to  the  line. 

The  projection  of  a  line-segment  on  a  given  line  in  the 
same  plane  is  the  portion  of  the  second  line  bounded  by 
the  projections  of  the  ends  of  the  first  line. 


Fig.  38. 

The  projection  of  AB  is   CD  in  I,  II,  and  III,  and  AD 
in  IV. 

B  D  % 


/ 

/N 

,  Z£ 

>> 

E 

D 

^^ 

V 

B 

a{- 

. r 

F 

a( 

/Jr 

V 

/ 

A' 

B' 

0'            Df 

E' 

A1 

D' 

a' 

B' 

Fig.  39. 

Fig.  40. 

The  projection  of  a  broken  line  is  the  sum  of  the  projec- 
tions of  its  parts. 

57 


58 


PLANE    TRIGONOMETRY 


The  projection  of  ABODE  (Fig.  39)  is  A'B'  -f  B'C  +  CD' 
4-  D'E'=A'E';  and  the  projection  of  ABCD  (Fig.  40)  is 
A'B' +  B'C  +  CD' =  A 'D'.     (See  §  19.) 

It  is  obvious  that  the  projection  of  a  broken  line  is  equal 
to  the  projection  of  the  straight  line  connecting  the  ends 
of  the  broken  line.  It  is  to  be  noted  that  here  we  take 
the  direction  of  the  lines  into  account.  The  projection  of 
ABCD  (Fig.  40)  is  equal  to  the  projection  of  AD,  and  is 
the  negative  of  the  projection  of  DA. 

38.(  The  projection  of  a  line-segment  on  any  line  in  its 
plane  is  equal,  in  length  and  direction,  to  the  length  of  the 
segment  multiplied  by  the  cosine  of  the  angle  which  the 
segment  makes  with  the  line.)  In  the  figure  the  line-seg- 
ment is  AC,  the  line  of  projection  is  LM,  and  the  angle, 
measured  according  to  §  20,  is  </>. 


The  projection  of  AC  is  A'C  =  AB. 


Now        cos  <f>  =  - 


AB 
AC 


\  AB  =  A'C  =  AC  cos<£. 


The  projection  is  positive  when  </>  is  an  angle  of  the  1st 
or  4th  quadrant ;  it  is  negative  when  <f>  is  an  angle  of  the 
2d  or  3d  quadrant. 


THE   ADDITION  FORMULA 


59 


39.  Projection  on  Coordinate  Axes.  The  proj  ections  of  A  C  on 
XX'  and  YY'  may  be  called  the  ^-projection  of  AC  and  the 
^/-projection  of  AC.  Let  <j>  be 
the  angle  which  A  C  makes  with 
XX'.     Draw  OB  parallel  to  AC. 

<£  =  XOD,     .'.  YOB  =  cf>-  90°. 

If  <£  is  less  than  90°,  as  in  the 
case  of  A'C,  the  angle  YOB'  is 
negative  and  equal  to  90°  —  </>. 

But    -  (90°  -<£)  =  <£-  90°. 

In  any  case  the  angle  which 
AC  makes  with  YY'  is  90°  less  than  the  angle  it  makes 
with  XX'. 

cc-projection  of  AC  =  AC  cos  <£. 
y-projection  of  AC  =  AC  cos(<£  —  90°)  . 

=  AC  cos(90°-4>)  by  [9] 

—  AC  sin  <£. 


40.  The  Addition  Formulas.  These  formulas  enable  us  to 
express  the  functions  of  the  sum  or  the  difference  of  two 
angles  in  terms  of  the  functions  of  the  constituent  angles. 
Without  examining  the  matter,  the  student  might  make  the 
mistake  of  writing : 

sin  (<f>  -f-  0)  =  sin  <£  +  sin  0. 
tan  (<£  +  0)  =  tan  cf>  +  tan  0,  etc. 

In  the  accompanying  figure  the 
points  F  and  C,  on  the  terminal  lines 
of  <f>  and  0,  respectively,  are  taken 
on  the  circumference  of  the  unit 
circle.     We  have,  therefore,  in  line 


Fig.  43. 


representatives  (see  §  29)  : 


60 


PLANE   TRIGONOMETRY 


sin  <j>  =  AF,  sin  6  =  EC,  sin  (<f>  +  0)=  BC. 
tsii<t>  =  LT,   tnnO=FD,  tan  (<£  +  0)  =  Z  Q. 

It  is  evident  that  AF  +  EC>  BC, 
or  sin  <j>  -f-  sin  0  >  sin  (<£  +  0), 

and  LT  +  FD<  LQ, 

or  tan  <£  +  tan  0  <  tan  (<£  + 0). 

Since  the  formulas  fail  in  this  particular  case,  they  are 
obviously  untrue. 

41.    The  Sine  and  Cosine  of  <|>  +  8. 


vAT 

Y 

<2\\ 

/M 

/X 

A 

0     ii 

N 

Fig.  44. 

Let  LOM=  <j>  and  MON  =  0,  then  LON  =  <f>  +  0. 

In  I,  <£  +  (9  <  90°;  in  II,  <£  +  (9  >  90°;  in  both,  <£  <  90°, 
0  <  90°. 

Through  P,  any  point  in  OM,  draw  TtPQ  perpendicular 
to  OM. 

Angle  LRQ  —  90  +  <t>,  being  the  exterior   angle  of  the 
triangle  OPtf. 

Since   OQ  is  a  line  connecting  the   extremities  of   the 
broken  line  OPQ,  we  have,  by  §  37  : 

^/-projection  of  OQ 

=  ^-projection  of  OP  -f-  ?/-proj  ection  of  PQ,     [12] 

x-proj  ection  of  OQ 

=  x-proj  ection  of  OP  -f  cc-proj  ection  of  PQ.     [13] 


U  -f  4  )  c   e^~  A^  X  -  - —    * 

v  Oy  THE   ADDITION   FORMULA        ^  61 

Applying  §  39,  these  equations  become: 

OQ  sin  (<£  +  0)  =  OP  sin  <j>  +  PQ  sin  (90°  -f  <£),    [14] 
6>Q  cos  (<£  +  0)  =  OP  cos  <t>  +  PQ,  cos  (90°  +  <£).    [15] 

But  OP  =  OQ  cos  0,  PQ  =  OQ  sin  (9. 

sin  (90°  +  <£)  =  cos  <£,  cos  (90°  +  <£)  =  -  sin  <£. 

Substituting  these  values  in  [14]  and  [15],  we  have 

OQ  sin  (4>  +  0)  =  OQ  sin  <£  cos  0  +  OQ  cos  <£  sin  (9,     [16] 

OQ  cos  (<£  -f-  0)  =  OQ  cos  <£  cos  0  -  OQ  sin  <£  sin  0.      [17] 

.*.  sin  (<j>  +  0)  =  sin  <£  cos  0  +  cos  <j>  sin  0,  [18] 

cos  (<£  -f  0)  =  cos  <£  cos  0  —  sin  <j>  sin  0,  [19] 

fir 

where  <£  and  0  are  both  less  than  90°  (  — 

42.    To  establish  the  truth  of  these  formulas  where  <f>  and 
0  are  unlimited  we  proceed  as  follows  : 

Let  <t>  =  90°  +  P,  where  0  <  90°. 
sin  (cj>  4-  0)  =  sin  (90°  4-0  4-0)=      cos  (0  4-  0),      [20] 
cos  (<£  4-  0)  =  cos  (90°  4-04-0)  =  -  sin  (0  4-  0).      [21] 

Since  0  and  0  are  each  less  than  90°, 
sin  (<£  4-  0)  =      cos  (04-0)=      cos  0  cos  0  -  sin  0  sin  0,  [22] 
cos  (<£  4-  0)  =  -  sin  (04-0)  =  -  sin  0  cos  0  -  cos  0  sin  0.  [23] 

But    sin  0  =  sin  (<£  -  90°)  =  -  sin  (90°  -<£)=-  cos  <£, 
cos  0  =  cos  (<£  -  90°)  =      cos  (90°  -  <£)  =       sin  <£. 

Substituting  these  values  in  [22]  and  [23], 

sin  (<£  +  0)  =  sin  <f>  cos  0  +  cos  <£  sin  0,  [18] 

cos  (<£  4-  0)  =  cos  $  cos  0  —  sin  <£  sin  0.  [19] 

Here  <£  is  an  angle  of  the  2d  quadrant,  0  an  angle  of  the 
1st  quadrant.     By  a  repetition  of  this  process  we  can  show 


62  PLANE   TRIGONOMETRY 

that  [18]  and  [19]  hold  when  <£  is  of  the  3d  quadrant,  etc. 
Treating  0  similarly,  we  prove  these  formulas  true  for  all 
positive  values  of  <j>  and  0. 

43.  They  also  hold  when  one  or  both  the  angles  are 
negative. 

Let  <f>  =  p  -  90°  where  p  <  90°. 
sin  (<£  +  0)  =  sin  (fi  -  90°  +  0) 

=  -  sin  (90°  -  f3  —  0)  =  -  cos  (p  +  0),  [24] 

cos  (<£  +  0)  =  cos  (p  -  90°  +  0) 

=      cos  (90°  -  p  -  0)  =      sin  (£4-0).  [_2^ 

Since,  p  and  0  are  positive, 
sin  (<£  +  0)  =  -  cos  (p  +  (9)  =  -  cos  0  cos  (9  +  sin  £  sin  (9,  [26] 
cos  (<£  +  0)  =      sin  (0  +  0)  =      sin  0  cos  0  4-  cos  0  sin  0.  [27] 
But  sin  p  =  sin  (<£  +  90°)  =  +  cos  <£, 

cos  p  =  cos  (<£  +  90°)  =  -  sin  </>. 
Substituting  these  values  in  [26]  and  [27], 

sin  (<£  -|-  0)  =  sin  </>  cos  0  4-  cos  <£  sin  0,  [18] 

cos  (<£  4-  0)  =  cos  <£  cos  0  —  sin  <£  sin  0.  [19] 

A  similar  process  of  reasoning  would  show  that  these  for- 
mulas remain  unchanged  when  both  <f>  and  0  are  negative. 
They  are  true  for  all  positive  and  negative  values  of  </>  and  0. 

These  formulas  are  so  important  that  they  should  be 
carefully  memorized.  They  may  be  translated  into  words 
as  follows  : 

I.  The  sine  of  the  sum  of  two  angles  is  equal  to  the  sine 
of  the  first  into  the  cosine  of  the  second,  plus  the  cosine  of 
the  first  into  the  sine  of  the  second. 

II.  The  cosine  of  the  sum  of  two  angles  is  equal  to  the 
product  of  their  cosines  minus  the  product  of  their  sines. 


THE   ADDITION  FORMULA 


63 


44.  The  sine  and  the  cosine  of  <£  —  0. 
Putting  -  0  for  0  in  [18]  and  [19],  we  have 

sin  (<£  —  0)  —  sin  <£  cos  (—  0)  +  cos  <£  sin  (—  0),     [28] 
cos  ((f)  —  0)=  cos  <f>  cos  (—  0)  —  sin  <j>  sin  (—  0).     [29] 

But  sin  (—  <f)    =  —  sin  <£,  cos  (—  <£)  =  cos  <$>. 

Substituting  these  values  in  [28]  and  [29],  we  have 

sin  (<f>  —  0)  =  sin  <f>  cos  0  —  cos  <£  sin  0,  [30] 

cos  (<£  —  0)=  cos  <£  cos  0  +  sin  <£  sin  0.  [31] 

Formulas  [18]  and  [30],  and  [19]  and  [31],  may  be  com- 
bined as  follows  : 

sin  (<f>  ±0)=  sin  <£  cos  0  ±  cos  <£  sin  0,  [32] 

cos  (4>±  6)=  cos  <£  cos  9  qp  sin  <£  sin  0.  [33] 

It  should  be  noted  that  in  [32]  the  double  sign  in  the 
second  member  is  like  the  double  sign  in  the  first  member, 
while  in  [33]  it  is  unlike. 

9. 

45.  Formulas  [18]  and  [19]  are  so  important  that  other 
geometrical  proofs  are  added. 


M 

zt 

XE 

(P 

J^\ 

Let 
then 


BAG 

I 

Fig.  45. 

LOM=  <j>  and  MON  =  MON'  =  0, 

LON  =  <£  +  0  and  LON1  =  <f>  -  Q. 


64  PLANE    TRIGONOMETRY 

Through  P,  any  point  in  OM,  draw  QPR  perpendicular  to 
OM.  Draw  PA,  QB,  and  RC  perpendicular  to  OL.  Draw 
PD  and  RE  parallel  to  OL. 

Z.DQP  =AEPR  =  <l>}  since  their  sides  are  perpendicular 
to  LO  and  OM. 

AP       DP       ER  OA       QD       EP 

sm  </>  — = = ?     cos  d>  = = = 

"       OP       QP       PR  *       OP       QP       PR 

A      PQ      PR  A       OP       OP 

sin  v  =  —  = 9  cos  6  =  —  =  —  • 

OQ       OR  OQ       OR 

sin(*  +  «)   -  — -- _- 

"  OQ        0Q~  OP'  OQ       QP'  OQ 

=  sin  <£  cos  0  +  cos  <£  sin  0.  [18] 

cos  (</>  +  0)    = 


[19] 


cos(<£  —  0) 


OQ             OQ 

OA       DP  _  OA 

OQ       OQ  ~  OP 

OP       DP 
OQ       QP' 

QP 
OQ 

cos  <£  cos  0  —  sin 

<£  sin  0. 

RC      AP  -  EP 

OR  ~         OR 

AP       EP  _  AP 
or"  OR"  OP 

OP       EP 
OR       PR 

PR 
OR 

sin  <£  cos  $  —  cos 

<j>  sin  0. 

<9C       OA  +  J577? 

0i2  ~         OR 

OA       ER  __  0-4 
OR       OR"  OP 

OP       ER 
OR       PR 

PR 
OR 

[30] 


cos  <£  cos  0  +  sin  <f>  sin  0.  [31] 


THE   ADDITION   FORMULA 


65 


46.    Still  another  proof  of  [18]  and  [19]  is  given  below. 

Construction.  Lay  off  OA  =  unity.  Draw  AB  and  AQ 
perpendicular  to  OM  and  ON,  respectively.  Draw  BC  per- 
pendicular to  ON.     Draw  AD  perpendicular  to  BC. 

Z  ABB  =  0  ;   their  sides  are  perpendicular. 
Let  sin  cj>,  cos  <£  =  su  cx, 

sin  0,  cos  0  =  s2,  c2. 


Fig.  46. 


The  lines  in  the  figure  evidently  have  the  lengths  indi- 
cated.    For  example,  BC  =  cxs2,  etc. 

AQ 

sin  (<f>  +  0)    =-—  =  AQ  =  CD  =  BD  +  CB  =  sxc2  +  cts2 
UA 


sin  cf>  cos  0  -f  cos  <f>  sin  0. 


47.    Tangent  and  Cotangent  of  <j>  +  9  and  ((>  —  0. 

sin  (<fr  +  0)  _  sin  <£  cos  0  -f  cos  <£  sin  $ 
cos  (<£  +  0)       cos  <£  cos  0  —  sin  <f>  sin  0 

Cf.  [4],  [18],  [19] 


tan(<£  +  0)  = 


66  PLANE   TRIGONOMETRY 

Dividing  both  numerator  and  denominator  by  cos  <f>  cos  0, 
we  have 

tan  d>  -f  tan  0  __  .  _ 

**«  +  9>=i-L*»r6  [34] 

Other  forms  for  tan  (<£  +  0)  may  be  found  by  dividing 
by  sin  <j>  sin  <£,  sin  <f>  cos  0,  cos  <j>  sin  0,  instead  of  cos  <f>  cos  0. 
Find  them.  Why  is  [34]  preferred  ?  In  like  manner  we 
find 

fato/j,       m        sin(<ft-fl)  _    tan<fr-tan0 

tan  (*-*)-  COs(</>-0)  ~  lTta^tanT  [3°] 

Note.  [35]  might  be  obtained  from  [34]  by  putting  —  </>  for  <p. 
Verify  this  statement. 

Similarly 

-  cos  (^  ~^"  ^)  _  cos  ^  cos  ^  Z  s^n  ^  E2  ^ 
'        sin  (<£  -f-  0)       sin  <£  cos  0  +  cos  </>  sin  0 

Dividing  numerator  and  denominator  by  sin  </>  sin  0,  we 
have 

.  ',  ,    ,    *v         cot</>  cot  0  —  1 

cot(<£  +  0)   = -  — — — •  [36] 

J  cot  0  -f-  cot  </> 

In  like  manner 

^     vv^  cot  0  -  cot  </> 

Find  other  forms  for  [36]  and  [37]  by  dividing  by  cos  <£ 
cos  0,  by  cos  <£  sin  0,  by  sin  <j>  cos  0,  instead  of  by  sin  <j>  sin  0. 


EXERCISES 

1.  Deduce  [36]  and  [37]  from  [34]  and  [35]  by  using  [3]. 

2.  Deduce  [36]  and  [37]  from  [34]  and  [35]  by  substi- 
tuting (90  +  <£)  for  cf>  in  the  latter. 

3.  Prove 

sin  (<£  +  0)  sin  (<£  —  #)  =  sin2<£  —  sin20  =  cos20  —  cos2<£. 


THE   ADDITION  FORMULA  67 

4.  Prove 

cos  (<£  -j-  0)  cos  (<£  —  0)=  cos2<£  —  sin20  =  cos20  —  sin2<£. 

5.  Find  formulas  for  sec (<f>  +  0),  sec (</>  —  0),  esc (<£  +  0), 
esc  (<£  —  0)  in  terms  of  the  secants  and  cosecants  of  <f>  and  0. 

6.  Find  the  sine  of  75°. 

sin  75°  =  sin  (45°  +  30°)  =  sin  45°  cos  30°  +  cos  45°  sin  30° 
=  iV2.£V3_+iV2.i  (p.  18) 
=  i(V6  +  V2). 

7.  Find  the  other  functions  of  75°. 

8.  Find  all  the  functions  of  15°.     (15°  =  45°  -  30°.) 

9.  Find  all  the  functions  of  180°.     (180°  =  90°  +  90°.) 

10.  Find  all  the  functions  of  135°.     (135°  =  90°  +  45°.) 

11.  sin  <£  =  ■£-,  sin  0  =  -J- ;  find  the  functions  of  <j>  +  0  and 
<j>-6. 

12.  Prove  sin_1#  +  sin_1?/=  sin-1  (x  Vl  —  y2  \-y  Vl  —  x2). 

13.  Prove  ^_ 

cos-1#  +  cos-1?/  =  cos-1  (xy  —  V(l  —  x2)  (1  —  y2)). 

14.  Prove  tan-1#  +  tan-1?/  =  tan" 


,  x  4-  v 

-1^/  —  fo-n  — 1  1£— 


1-xy 

48.    Functions  of  Double  Angles.     If  we  put  0  =  <fy  in  for- 
mulas [24],  [25],  [34],  and  [36],  we  shall  have 

sin  (<£  +  <£)=  sin  <£  cos  <£  -+-  cos  </>  sin  <£. 

. ' .  sin  2  <£  =  2  sin  <£  cos  <£.  [38] 

cos  (<£  +  <£)  =  cos  <£  cos  <£  —  sin  <j>  sin  <£. 
cos  2  <£  =  cos2<£  —  sin2<£  I  ^| 

=  2  cos2<£  -  1,    since  sin2<£  =  1  -  cos2<£      II  [  [39] 
=  1  -  2  sin2<£,        "     cos2<£  =  1  -  sin2<£.   Ill . 


68  PLANE   TRIGONOMETRY 


EXERCISES 

1.  Given  the  functions  of  30°,  find  those  of  60°,  of  120°, 
of  240°. 

2.  Given  the  functions  of  45°,  find  those  of  90°,  of  180°, 
of  360°. 

Prove  the  following : 

2  —  sec2^c 

3.    = =  cos  2  x.  4.    tan  x  -f-  cot  x  =  2  esc  2  x. 

sec2# 

5.  (sin  x  ±  cos  a?)2  =  1  ±  sin  2  #. 

2  tan  x  .    A       ^^...^  1  +  tan2# 

6.  — — — —  =  sm2a;.   Cf.  [401.     7.   --L— — —  =  sec  2  x. 
1  +  tan2ic  J   L     J  1  -  tan2x 

8.  Find  formulas  for  sec  2  0  and  esc  2  0. 

9.  2  sin  (45°  -f  </>)  sin  (45°  -  <£)  =  «**$£    C^3L  ^  $> 


49.    Functions  of  Half-angles.     If  in  III  and  II  of  [39] 
we  substitute  \  <f>  for  <$>, 


cos  <£  =  1  —  2  sin2  -^  <£. 
cos  <£  =  2  cos2  ^  <£  —  1. 

•     i              <*/l-~  cos  <£ 

[42] 

, '          ,  .  /l  +  cos  <f> 
cos£<£  =  ±\               r- 

[43] 

THE   ADDITION  FORMULA  69 

By  formula  [4] 


t    ,  1  —  COS  <f> 

*  1  +  cos  <f> 

_  1  —  cos  <£ 
sin  <j> 
sin  <£ 


II 


III 


[44] 


1  +  cos  <j> 

II  is  derived   from   I    by  multiplying    numerator  and 
denominator  by  1  —  cos  <£ ;  while 

III  is  derived  from  I  by  using  1  +  cos  <f>  as  multiplier. 
By  formula  [3] 


4.    1     A  ^  l1    +    C0S   <t> 

cot^-<ft  =\ 


COS  <j> 

sin  <£ 


1  —  cos  (j> 

1  -f  cos  <f> 

sin  <£ 


II 
III 


[45] 


EXERCISES 

1.  Given  the  functions  of  60°,  find  those  of  30°,  of  15°. 

2.  Given  the  functions  of  45°,  find  those  of  22°  30'. 

3.  Given  sin  <£  =  £,  find  the  functions  of  -t^- 

4.  cos  <£  =  a* ;  find  the  functions  of  ^* 
Verify  the  following : 

1  +  sec  <£  <£ 

5.   -— -  =  2cos2tt- 

sec  <£  2 

6.  cos2|(  1  +  tan|  J  =  1  +  sin  <£. 

7.  esc  as  —  cot  x  =  tan  -• 

8.  sin2  -  (  cot  -  —  1  ]  =  1  —  sin  x. 


/ 


y 


70  PLANE   TRIGONOMETRY 

50.    Functions  of  Three  Angles. 

sin(K+^V§=sin(^+f+y)  u         ^ 

=  sin  (a  +  ft)  cos  y  +  cos  (a  +  /3)  sin  y 

=  (sin  a  cos  ft  +  cos  a  sin  ft)  cos 

+  (cos  ^  cos  'ft  —  sin  a  sin  2) 

=  sin  a  cos  /?  cos  y  +  cos  er  sin  5  cos  y 

+  cos  a  cos  /3  sin  y  —  sin  a  sin  ft  sin  y.  [46] 

Similarly  " 

cos  (a  +  ft  -{-  y)  =  cos  a  cos  ft  cos  y  —  sin  a  sin  /3  cos  y 

—  sin  a  cos  ft  sin  y  —  cos  a  sin  /3  sin  y.  [47] 

tan  a -h  tan /3  + taiiy  —  tanatantftany  _JA_ 

tan  (a  +  0  +  y)  =  =— — — -^ -— *— — ^- — £•  [48] 

\      ^      '7      1— tan /Stan  y—  tan  y  tan  a— tanatan/2  L     J 

Formula  [48]  may  also  be  obtained  by  dividing  [46]  by 
[47],  and  then  dividing  numerator  and  denominator  by  cos  a 
cos  ft  cos  y. 

If  now  in  [46],  [47],  and  [48]  we  put  ft  =  y  =  a,  we 
shall  have 

sin  3  a  =  3  cos2a  sin  a  —  sin3a 

=  3  sin  a  —  4  sin8a.  [49] 

cos  3  a  =  cos3a  —  3  sin2a  cos  a 

=  4  cos3a  —  3  cos  a.  [50] 

0  3  tan  a  —  tan3a  _  M  - 

tan3a=     ^g^    •  [«t] 

EXERCISES 

1.  Put  the  last  six  formulas  into  words. 

2.  Find  the  sine,  the  cosine,  and  the  tangent  of  a  +  ft  —  y, 
of  a  —  ft  —  y. 

3.  Find  the  cotangent  of  a  -\-  ft  +  y  in  terms  of  the 
cotangents  of  the  constituent  angles. 


THE   ADDITION   FORMULA  71 

4;   Find  sin  3  <£  by  developing  sin  (2  <£  -f  <£)  by  formula 
[18]  and  simplifying  the  result. 

5.  Find  sin  4  <£  by  putting  2  <£  f or  <£  in  [38]. 

6.  Deduce  formula  [47]  from  [46]  by  putting  90  +  a 
for  a. 

7.  Find  cos  4  <£. 

8.  Find  sin  5  <£  and  cos  5  <£. 

9.  Given  the  functions  of  30°,  find  those  of  90°.      / 

51.    Conversion  Formulas.    By  adding  and  subtracting  [18] 
and  [30],  and  [19]  and  [31],  we  have 

sin  (<f>  +  0)  +  sin  (<£  —  0)  =      2  sin  <£  cos  0. 

sin  (<£  -f-  0)  —  sin  (<£  —  0)  —      2  cos  <£  sin  0. 

cos  (<£  -f-  0)  +  cos  (<£  —  0)  =      2  cos  <£  cos  0. 

cos  (<£  +  0)  —  cos  ($  —  0)  =  —  2  sin  <£  sin  0. 
Putting     (<£  +  0)  =  a,  (<£  -  0)  =  p, 
whence  <£  =  ^  (a  +  /?),  0  =  £  (a  —  ■/?),  we  have 

sin  a  +  sin  f3  =  2  sin  £  (a  +  /?)  cos  £  (a  —  /?),  [52] 
sin  a  —  sin  /J  =  2  cos  £  (a  +  /J)  sin  -J-  (a  —  /?),  [53] 
cos  a  +  cos  /?  =  2  cos  £  (a  +  /?)  cos  -J-  (a  —  /?),  [54] 
cos  a  —  cos  /3  =  —  2  sin  -J-  (a  -f-  /3)  sin  £  (a  —  /?).    [55^\ 

,  These  formulas  enable  us   to  express  the  sum  or  the 
difference  of  two  sines  or  two  cosines  as  a  product. 

EXERCISES 

1.  Express  the  last  four  formulas  in  words. 
Verify  these  formulas  when 

2.  a  =  60°,  p  =  30°.  4.    a  =  180°,  £  =    90°. 

3.  a  =  90°,  /?  =  60°.  5.    a  =  270°,  (3  =  180°. 


72  PLANE   TRIGONOMETRY 


Verify  the  following  identities  : 

sin  a  +  sin  ft  _  tan  %(a-\-  ft) 
..-   '    sin  a  —  sin  ft      tan  %(a  —  f3) 

sin  a  +  sin  ft 

7.    ■ ^  =  tan  £  (a  4-/3). 

^     cos  a  +  cos  $  " v        ^y 

cos  a  +  cos  5  .  , 

J'    cosa-cosg  =  -COt^a  +  ^COt^a-^ 

9.  sin  60°  +  sin  30°  =  2  sin  45°  cos  15°. 

10.  sin  40°  -  sin  10°  =  2  cos  25°  sin  15°. 

11.  cos  75°  +  cos  15°  =  2  cos  45°  cos  30°. 

12.  sin  5  x  +  sin  3  x  =  2  sin  4  cc  cos  x. 

sin  3  as  4-  sin  2  #  x 

13.   5 —  =  cot-- 

cos  2x  —  cos  3x  2 

14.  cos  (60°  +  x)  +  cos  (60°  -  a)  =  cos  a;. 

15.  tan  50°  +  cot  50°  =  2  sec  10°. 


16.  sin  2  cos_1#  =  2a;  Vl  —  x2. 

17.  cos  2  sin-1#  =  1  —  2x2. 

18.  cos  2  cos-1#  =  2  x2  —  1. 

19.  tan  2  tan_1#  = 


1  —  x' 


20.  tan_1ir  4-  tan-1?/  =  tan-1 —  •    (Take  the  tangent 

of  both  members.) 

IT 

21.  sin-1#  4-  cos_1#  =  —  • 

22.  sin-1  a  4-  cos-1?/  =  sin-1  (xy  4- V(l  —  x2)  (1  —  y2)). 

23.  tan-1^  4-  tan-1 -J-  =  ->or  —  •     Cf.  example  20. 
.    24.  tan"1!  =  tan-1!  4- tan"1!. 

7T 

.'.  y  =  tan-1!  4-  tan-1!  4-  tan-1  J. 


THE   ADDITION   FORMULA  73 

52.  Trigonometric  Equations.  Trigonometric  equations 
are  generally  best  solved  by  expressing  all  the  functions 
involved  in  terms  of  some  one  function  and  solving  the 
resulting  equation. 

Illustrations.     1.  sin  cf>  =  tan  cj>. 

sin  <£  .     -  A  1     \ 

sin  d>  = 7  •     .  .  sin  d>    1 =  0. 

cos  (j>  \  COS  $J 

.'.  sin  <£  =  0;      and      cos  <£  =  1. 

<£  =  0  and  7r,  <j>  —  0. 

The  solutions  are  therefore  <f>  =  0,  7r. 


2.    tan  <£  =  esc  <£. 
sin  <£  1 


sin2<£  a=  cos  <£,     1  —  cos2<£  =  cos  <f>, 


cos  <£      sm  <j> 

cos2  <j)  +  cos  <f>  =  l, 

cos  <£  =  £(—  1±V5), 

^cos'^-liVo"); 
but  since  £(—  1  -Vo)  is  numerically  greater  than  unity, 
this  solution  is  impossible ;  and 

^  =  cos-1i(-l+V5). 

3.    sin  0  +  cos  0  =  1. 

sin  0  +Vl  -  sin2  (9  =  1. 

l-sin2(9=(l-sin0)2. 
(1  -  sin  Of  -  (1  -  sin20)  =  0. 
(1  -  sin  0)  (1  -  sin  $  — •  1  -  sin  (9)  ==  0. 
(1  -  sin  (9)  (-  2  sin  0)  =  0. 
.'.  sin  0  =  1  and  0. 

.  6  =  |,    0,    7T. 

The  solution  0  q=  7r  does  not  satisfy  the  original  equation. 


74  PLANE   TRIGONOMETRY 

EXERCISES 

Find  the  values  of  <f>  that  satisfy  each  of  the  following 
equations : 

1.  cos  2  <£  -f-  cos  <j>  =  0.  6.  tan  <f>  -f  cot  <j>  =  2£. 

2.  tan  <£  =  n  cot  <j>.  7.  cot  0  =  2  cos  0. 

3.  sec  </>  —  tan  <£  ==  cos  <£.  8.  tan  <£  -f-  cot  <£  =  m. 

4.  sin  0  -f-  cos  <f>  =  tan  <£.  9.  tan  cj>  +  sec  <f>  =  a. 

5.  3sin0  +  4  cos  0  =  5. 

10.  If  sin  0  +  cos  (9  =  a,  then  sin  2  0  =  a2  -  1. 

11.  Z  cos  0  +  m  sin  0  =  0,  find  tan  -• 

MISCELLANEOUS   EXERCISES 

1.  From  sin  30°  =  cos  60°  =  \%  cos  30°  =  sin  60°  =  \  V3, 
sin  45°  =  cos  45°  =  \  V2,  find  all  the  functions  of  15°,  of 
75°,  of  105°. 

2.  From  sin  a  =  f ,  sin  /?  =  f ,  find  all  the  functions  of  • 
a  +  fi  and  a  —  f3. 

Prove  the  following : 

n     sin  (a  +  b)  +  sin  (a  —  6) 

3.   ) — — ^- ) -f  =  tan<x. 

cos  (a  +  o)-\-  cos  (&  —  6) 

sin  (a  dob)  . 

4.    * '-  =  tan  a  ±  tan  #. 

cos  a  cos  6 

c     cos  (a  qp  ft)  .       ,    ,       7 

5.  -t— ^ — =!—t  =  cot  a  ±  tan  6. 
sin  a  cos  & 

tan  #  +  tan  ?/  _  sin  (x  +  ?/) 
~-~-^*~"  tan  x  —  tan  ?/      sin  (x  —  y) 


7. 


1  —  tan  x  tan  ?/  _  cos  (x  4-  ?/) 
1  4-  tan  x  tan  y      cos  (x  —  y) 


_.     cot  ?/  +  cot  a?  ,  s        ,     .     . 

8.    — r2 ; —  =  esc  (x  —  y)  sin  (x  +  y). 

coty  —  cotx  \    ■     */        \    ■     &/ 


&ts 


THE   ADDITION  FORMULA 


75 


10. 
11. 

12. 
13. 
14. 
15. 

16. 

17. 

18. 
19. 
20. 
21. 
22. 

23. 
24. 

25. 
26. 

27. 


tan  x  cot  y  4- 1  _  sin  (#  +  y) 
tan  x  cot  y  —  1      sin  (as  —  y) 

sin  (as  +  y)  sin  (as  —  y) 


cos2as  cos2y 


tan2as  —  tan2?/. 


tanas. 


tan2as  —  tan2?/  ■      ,     .  ,       . 

— — - — if-  =  tan  (as  +  y)  tan  (a:  —  y). 

1  —  tan2cc  tan2y  v        VJ        K        VJ 

V  2  sin  (a  ±  45°)  =  sin  a  ±  cos  a. 

sin  (as  +  3/)  cos  x  —  cos  (as  +  y)  sin  as  =  sin  y. 

sin  (as  —  y)  cos  y  4-  cos  (as  —  y)  sin  ?/  =  sin  as. 

cos  (as  +  y)  cos  as  +  sin  (as  +  y)  sin  as  =  cos  y. 

tan  (a?  —  y)  4-  tan  y   _     tan  (x  +  y)—  tan  ?/ 

1  —  tan  (as  —  y)  tan  y      1  +  tan  (as  +  y)  tan  y 

2  sin  (45°  +  a)  cos  (45°  -  ft)  =  cos  (a  -  ft)  4-  sin  (a  4-  ft). 
(7/*.  exercise  12. 

2  sin  (45°  -  a)  cos  (45°  +  b)  =  cos  (a  -  ft)  -  sin  (a  4-  ft). 

2  sin  (45°  4-  a)  cos  (45°  4-  b)  =  cos  (a  4-  ft)  4-  sin  (a  -  ft). 

2  sin  (45°  —  a)  cos  (45°  —  ft)  =  cos  (a  4-  ft)  —  sin  (a  —  ft). 

tan  as  =  £,  tan  y  =  i ;  find  tan  (as  4-  y)  and  tan  (x  —  y). 

tan  as  =  3,  tan  y  =  -J-;  find  tan  (as  -f  ?/)  and  tan  (as  —  ?/). 

tan  as  =  k,  tan  y  =  t}    find,  cot  (as  4-  y)  and  cot  (as  —  y). 

rC 

i,     ,  ^on      cot  as  — 1      ^/l  — sin2as       1  — sin2as 
cot  (as 4-45  )  =  -— —7  =  \/t— — r-3-  =  — 

nnt.  *  4-1  M  4"  Sin  2  05 


cot  as  4-  1  14-  sin  2as 

cot  (a;  -  45°)  =  cotx  +  1  =  tan^  +  1. 
v  }      \  —  cot  as      tan  as  —  1 

tan  (as  ^  45°)  +  cot  (a  ±  45°)  =  0. 


cos  2  as 


tanas  = 


m  4- 1 


cot  y  =  2  m  +  1 ; 

find  tan  (as  4-  y)  and  cot  (as  —  y). 


76  PLANE   TRIGONOMETRY 

28.  If  x  +  y  +  z  =  90°,  then  tans  =  1~t^xt^y 

tan  x  -f-  tan  3/ 

29.  sin  7  #  —  sin  5  a;  =  2  sin  #  cos  6  x. 

30.  cos  5  #  +  cos  9  x  =  2  cos  7  #  cos  2  #. 

31.  cos #  —  cos  2x  =  2  sin  | x  sin  £ x. 

sin  2  03  —  sin  03 

32.    —  =  cot  I  x. 

cos  ^  —  cos  2x  J 

m     sin  3 x  —  sin  2x  ,  5x 

33. — -=*COt  — 

cos  2  03  —  cos  3  03  2 


34. 


sin  a?  +  sin  y  _  cos  #  +  cos  y 
cos  x  —  cos  y      sin  2/  —  sin  x 


35.  cos  (  -  —  x  j  —  cos  (  —  +  cc  j  =  sin  x. 

36.  cos  f  —  +  x  j  -f-  cos  f  —  —  x  )  =  cos  03.V2. 

Express  each  of  the  following  prodncts  as  the  sum  or 
difference  of  two  trigonometric  functions : 

37.  2  sin  x  cos  y.  40.    2  sin  3  03  cos  5  x. 

38.  2  cos  x  cos  2/.  41.    2  cos  (x  +  3/)  cos  (x  —  y). 

39.  2  sin  2  a;  cos  3  y.  42.    2  cos  f  x  cos  -J-  as. 

43.  2  sin  50°  cos  10°. 

aa        O  7T      .        7T 

44.  2  cos  —  sm— • 

4        12 

Simplify : 

45.  2  cos  3  x  cos  03  —  2  sin  4  x  sin  2  #. 
cos  x  —  cos  5  as 


46. 

47. 
48. 


sin  x  -f-  sin  5  03 

sin  3  x  —  sin  03      sin  3  x  —  sin  x 
cos  3  03  +  cos  03       cos  3  x  —  cos  03 
(sin  4  03  —  sin  2  03)  (cos  x  —  cos  3  x) 
(cos  4  03  +  cos  2  03)  (sin  03  +  sin  3  x) 


49 


THE   ADDITION   FOKMULA  77 


Verify  : 

cos  x  +  cos  3  x   __  cos  2  x 
cos  3  x  +  cos  5  a;      cos  4  a; 

#  +  ?/  x  —  y  2  sin  y 

50.  tan      n  J  —  tan  —7-^  = — - 

2  2  cos  a?  +  cos  y 

51.  2  sin  2  x  cos  a;  +  2  cos  4  a;  sin  cc  =  sin  5x  +  sin  a;. 

Lob    X  _ 

52.  — r -  =  sec2a;. 

csc2a;  —  2 

53.  cos2  a;  (1  —  tan2  x)  —  cos  2  a;. 

54.  cot  a;  —  tan  x  =  2  cot  2  a;. 
cos  2  x  1  —  tan  a; 


55. 


1  4-  sin  2  a;       1  +  tan  x 


56.  cos2a;  +  cos2(  ■«  +*  )  +  cos2(7r-f-#)-bcos2(  —+x  1=2. 

57.  sin  x  +  sin  3  a;  -f  sin  5  a?  +  sin  7  a;  =  4  sin  4  a;  cos  2  a;  cos  a;. 

58.  cos  x  4-  cos  (120  +  x)  4-  cos  (120°  —  a?)  =0. 

sin  3  x      cos  3  a; 

59.  — ; =  2. 


sin  x 

COS  X 

cos  3  a; 

4- 

sin  3  a? 

sin  x 

COS  X 

sin  5  a; 

cos  5  a; 

60.   — : 1 =  2  cot  2  x. 


61.  — =  4  cos  2  x. 

sm  a;  cos  x 

62.  tan  a;  =  I,  tan  ?/  =  T2T  ;  ■  find  tan  (2  x  4-  y). 

63.  sin  (y  -\-  z  —  x)-\-  sin  (z  4-  x  —  y)  4-  sin  (a;  +  y  —  z)  — 

sin  (a;  +  y  4-  2)  =  4  sin  a;  sin  y  sin  z. 

64.  cos  (?/  4-  z  —  x)  4-  cos  (z  +  x  —  y)-\-  cos  (a;  4-  y  —  z)  4- 

cos  (as  +  ?/  +  «)  =  4  cos  a;  cos  ?/  cos  z. 

65.  sin  a;  sin  (y  —  z)  4-  sini/sin(^  —  a;) -{-sin  2  sin  (a;  —  y)  =  0» 

66.  cos  a;  s in  (?/  — 2)  + cos  2/ sin  (2  —  x)  4-  cos  2  sin (x  —y)  =  0. 


78  PLANE   TRIGONOMETRY 

€7.    cos  a?  cos (^/  —  z)—  sin  7/ sin (^  —  x)  —  cos£COs(#  —  y)  =  0. 
68.    sin  x  cos  (y  —  z)  +  cosy  sin  (2  —  #)  —  sin^cos(x  —  y)  =  0. 

'X*  -  y*  4  jN 


69.    cot- *  (aj  —  y)  —  cot~ 1  (a?  +  y)  =  cot- *  ^ 


70. 

tan-1 —  tan-1 =  tan-1  ~-^- 

a  —  1                     a                     la1 

71. 

0    •      1         i        ^Wl-a2 

2  sin   x  a  =  tan-1  — — - — 

1  —  2  a2 

72.  tan-1*,  +  2  tan-1  ft  =  tan"1  "(1    7P  +  2  *. 

1  —  ft2  —  2  aft 

,1         .      ,  a  4-Va2-  1 

73.  tan_1a  4-  cos-1  -=  sin-1 . 

a  a  Va2  4  1 

74.  cos4  #  —  sin4  x  =  cos  2  #. 

75.  sin2  (x  +  y)  —  sin2  (x  —  y)=  sin  2  ^  sin  2  y. 

,76.    (sin  x  —  sin  y)2  +  (cos  x  —  cos  y)2  =  4  sin2  — r-^- 

1  4-  sin  .r  —  cos  x 

77. — ; =  tant^. 

1  4-  sm  x  4  cos  x 

tan  x  ±  tan  7        .   , 

78.    ; -  =  ±  tan  x  tan  y. 

cot  x  ±  cot  y 

79.  (sin  <£  4  sin  0) (sin  <£  —  sin  0)  =  sin  (<£  +  0)  sin(<£  —  0). 


80.    (Vl  4  sin  a  —  Vl  -  sin  a)2  =  4  sin2|a. 


81.    (Vl  4  sina+Vl  —  sina)2=  4  cos2  J  a. 

2 


82.  sec  2  a  +  tan  2  a  +  1  = 

83.  sin  A  = 
,  84.    cot  (x  +  y)  = 


1  —  tan  a 
sin  (30°  +  A)-  sin  (30°  -  A) 
V3 
1  1 


tan  x  4-  tan  y      cot  #  4  cot  y 


CHAPTER  VI 
THE   TRIANGLE 

53.  The  object  of  this  chapter  is  to  study  the  relations 
between  the  sides  of  a  triangle  and  the  trigonometric  func- 
tions of  its  angles.  Other  properties  of  the  triangle  are 
also  considered. 

NOTATION 

A,  B,  C  =  the  vertices  of  the  triangle. 
a,b,c  =  the  sides  opposite  A,  B,  C,  respectively, 
a,  ft,  y  =  the  interior  angles  at  A,  B,  C,  respectively. 
s  =  %  (a  +  h  +  c),  the  semi-perimeter. 
B,  r  =  the  radii  of  the  circumscribed  and  inscribed 
circles. 
ra>  rhi  rc  s  the  radii  of  the  escribed  circles  opposite  A,  B,  C, 
respectively. 
jpa,  pb,  pc  =  the  altitudes  from  A,  B,  C  to  a,  b,  c. 
K  =  area  of  the  triangle. 

MEMORANDA 

a  +  p  +  y  =  180°  =  tt. 

a,p,y  =  7r-(p  +  y),7r-(y  +  a),7r-(a  +  f3). 

sin  a,  sin  /?,  sin  y  =  sin  (/J  +  y),  sin  (y  +  a),  sin  (a  +  /?). 

cos  a,  cos  /?,  cos  y  =  —  cos  (/?  +  y),  —  cos  (y  +  a),  —  cos  (a  +  /?). 

79 


80 


PLANE   TRIGONOMETRY 


tan  a  =  —  tan  (/?  +  y),  etc. 
cot  a  =  —  cot  (f3  -+-  y),  etc. 
sin  £  a  =      cos  %(fi  +  y),  etc. 
cos  i  a  =  +  sin  %({$  +  y),  etc. 
tan  -J-  a  =  +  cot  £  (/?  +  y),  etc. 
cot  i  a  =  4-  tan  £  (/?  -f  y),  etc. 
2  JT  =  apa  =  bph  =  cpc  =  2  rs. 
b  -f-  c  —  a  =  2  (s  —  a), 
c  +  a  -b  =  2(s  —b). 
a  4-  b  —  c  =  2  (s  —  c). 

54.    The  Law  of  Sines. 

In  either  figure,  let  AE  =  q,  then,  §  19,  EB  =  AB  —  AE. 


Fig.  47. 


sin  a  =  =^7         sin/3  =  — • 
b  a 

sin  a  _ffc      £c_« 


sin/3 
This  may  be  written 


Similarly 


sin  a  sin  /? 

b  c 

sin/3  siny 

a  b  c 

sin  a  sin/3       siny 


[56] 


THE   TRIANGLE  81 

This  is  the  law  of  sines.  It  may  be  stated  in  words  as 
follows :  The  ratio  of  the  side  of  any  triangle  to  the  sine 
of  its  opposite  angle  is  constant. 

Let  ns  denote  this  constant  by  M.     The  formula  becomes 

JL.=-J-  =  -J^  =  M.  [57] 

sin  a       sm  ft      sin  y  L     J 

55.  The  Law  of  Tangents.      From  [57], 

aba       sin  a 

t  or 


sin  a       sin  ft  b       sin  ft 

by  composition  and  division 
a  +  b  _  sin  a  +  sin  ft 
a  —  b      sin  a  —  sin  ft 

2sm|(tt+£)cosi(tt-£)         [52]and[53] 
2  cos  £  (a  +  ft)  sin  £  (a  —  ft)       J  L     J         L     J 

^tanj(a+ft) 
tan  i  (a  —  ft) 
§   a  +  5  =  tan  fr  (a  +  ft)  =        cotjy'      ^ 
''a-  b      tallica- ft)       tan£(a-ft)'  L     J 

If  b  is  greater  than  a  we  can  avoid  negative  signs  by 
writing  b  —  a  and  /3  —  a  instead  of  a  —  b  and  a  —  ft. 

Similar  formulas  may  be  derived  involving  b  and  c,  and 
c  and  a. 

This  is  the  law  of  tangents.  In  words  it  is :  The  ratio 
of  the  sum  of  any  two  sides  of  a  triangle  to  their  differ- 
ence is  equal  to  the  ratio  of  the  tangent  of  one-half  the 
sum  of  the  opposite  angles  to  the  tangent  of  one-half 
their  difference. 

56.    The  Law  of  Cosines.     From  Fig.  47, 

a2  =  (c  —  9.Y  +  V?,  Pc*  =  b2  —  q2. 
.-.    a2  =  (c  -  q)2  -\-b2-q2  =  b2-{-C2-2  cq. 


82  PLANE    TRIGONOMETRY 

But  q  is  the  projection  of  b  and,  therefore, 

q  =  b  cos  a. 
Substituting  in  the  preceding  equation, 
a2  =  b2  +  c2  —  2  be  cos  a. ' 


Similarly         b2  =  c 2  -f  a2  —  2  ca  cos  /3.  i- 

[59] 

c2  =  a2  +  b2  -  2  ab  cosy.  J 

Solving  for  cos  a,  etc.,                                     v 

62  +  c2-a2   • 
cos  a  = — ■ 

2&c 

fa 

[60] 

This  is  the  Zaw  0/  cosines. 

57.    Functions  of  the  Half -angles  in  Terms  of  the  Sides. 
Substituting  %  a  for  <£  in  [39]  III, 

cos  a  =  1  —  2  sin2  \  a. 
2  sin2  \  a  ==  1  —  cos  a 

by  [60] 


1- 

b2  +  c2- 

a2 

2  be 

a2. 

-{b2 

■Ye2- 

-2  be) 

2  be 

a2- 

-(b 

-e)2 

2  be 
(a  +  b  —  c)  (a  —  b  -f  c) 
2be 


-s  A — J_i 1.  (See Memoranda.) 


Similarly 


2  be 

171 

* .  sin  ^  a 

sin£/2 
sin  |  y 


=V(- 

be 

[61] 

-# 
-# 

-  c)  (s  —  a) 

ca 
-a)(s-  b) 

ab 

[61] 
[61] 

THE   TRIANGLE 


83 


Substituting  \  a  f or  <£  in  [39]  II, 

cos  a  =  2  cos2  \  a  —  1. 
2  cos2 |a  =  l  +  cos  a 

+         2fo 

_  2  5c  +  62  +  c2  -  a1 

"  26c 

_   g  +  c)2  -  ^2 


2fo 

""  2  6c 

= b -•    (See  Memoranda.) 

2  be  v  ' 


/S  (5  —  ») 

Is  (s  -  ^ 
008  i") 


ca 


*V-^- 


[62] 


Dividing   [61]  by  [62], 


smja 

=  tan  \  a 
cot -J- a 

\(s 

-b)(s-c) 

cos^-a 

—  \ 

s(s  —  a) 

=v. 

s(s  —  a) 

-b)(s-cy] 

l[63] 


Since  %  a,  £  /?,  -J-  y  <  90°,  the  functions  of  these  angles 
are  positive  and  the  radicals  in  [61],  [62],  [63]  are  also 
positive. 


84  PLANE   TRIGONOMETRY 

EXERCISES 

Verify  the  following  relations  : 

s  _  cos  \  /3  cos  ^  y  s  —  a  _  cos  ^  a  sin  -J-  y 

a  sin  \a  b  cos  ^  /3 

s  —  a  ^  sin  j-  ft  sin  j-  y 
a  sin  ^-  a 

4.  cos  a  +  cos  /?  cos  y  =  sin  ft  sin  y. 

5.  a  cos  ft  +  b  cos  a  ==  c. 
c  cos  a  +  &  cos  y  =  b. 
c  cos  ft  +  &  cos  y  =  a. 

„       ,  a2  -  62 

6.  a  cos  B  —  b  cos  a  = 

c 

7.  a  cos  ft  cos  y  +  £  cos  y  cos  a  -|-  c  cos  a  cos  ft 

=  \  \a  cos  a  +  6  cos  ft  -f-  c  cos  y] 

=  a  sin  ft  sin  y  =  b  sin  y  sin  a  =  c  sin  a  sin  ft. 

8.  a  sin  (ft  —  y)  -f-  b  sin  (y  —  a)  +  c  sin  (a  —  ft)  =  0. 

58.    Circumscribed  and  Inscribed  Circles. 

Circumscribe  the  circle  0  about  the  triangle  ABC.    Draw 
CD,  a  diameter.     Angle  a  =  angle  D.     (Fig.  48.) 

sina  =  sinD  =  ^  =  ^. 

.'.  2  i?  =  -A-  =  -A-  =  -A-  =  M.        [64] 
sin  a       sm  ft      sm  y 

Inscribe  the  circle  0  in  the  triangle  ABC.     By  geometry 

&i  =  C*  h  =  a2>  ci  =  K     (Fig-  49.) 
.-.  s  =  %(a  +  b  +  c)=  ax  +  bl  +  cx  =  ax  +  a2  +.«,  =  a  +  c^ 
.*.^4JF  =  (Si  =  5  —  a. 


THE   TRIANGLE 


85 


Now 


Similarly 


Fig.  48. 
Combining  [63]  and  [65^\, 


c,  F       c2    £\ 

Fig.  49. 


~  ft)  («  ~  <0 


s(s  —  a) 


..  ,.     J(* -<*)(*-•*)(*- c) 


[66] 

o 

From  [65]  we  have 
r  =  (s  —  a)  tan  £  a  =  (5  —  b)  tan  k  P  =  (s  —  c)  tan  J  y.    [67] 

Let  0  be  escribed  to  the 
triangle  ABC  opposite  A. 
We  have  by  geometry 

BD  ==  J3F,  CD  =  C% 
0£  =  BF  +  C#. 
.'.  2s  =  AE  +  AF, 
s  =  AE. 


tania  =  — -~  =  — 
2  ^4£        5 


[68] 


Similarly 


tan  -J  /?,   tan  |-  y  = 


86  PLANE   TRIGONOMETRY 


Combining  [63]  and  [68], 


ra  =  J(s  ~b)(s~  c) 
s        *       s  (s  —  a) 


.:ra,rh>rc=^(S-b^S-C),et,        [69] 
s       a 

Comparing  [65]  and  [68], 

rs  =  ra(s  —  a)  =  rb(s  —  b)  =  rc(s  —  c).  [70] 

EXERCISES 

Verify  the  following  identities  :  ^ 

,  ,  o  ara  +  hrb  +  CTc 

1-    ra+rb  +  rc  —  3r  = - 

11        1,1 

2.  -  =  -  +  -  +  -• 
r      ra      rb      rc 

r   —  v 

3.  tan  i  a  =  — 

2  a 

4.  OOx  =  (ra  —  r)  esc  %  a. 

5.  OOx  =  a  sec  ^  a  =  b  sec  -J-  ft  =  c  sec  \  y. 

v 

6.  tan-J-a  tan-J-jStan^y  = -• 

7.  sina  +  sin/2  +  sin  y  =  — • 

R 

/       5£.    Area  of  the  Triangle. 

/     •    The  area  of  a  triangle  may  be  expressed  in  different 
ways,  depending  upon  the  parts  known.     We  have 
geometry  (Tig.  51) 

2K=apa  =  bpb  =  cpc, 
pa  =  c  sin  (3  =  b  sin  y. 
.\  2  K  =  ac  sin  f3  =  ab  sin  y  =  be  sin  a. 


THE   TRIANGLE 


87 


a  smy 

c  =  — : '  - 

sin  a 


From  [57], 

Substituting  this  value  in  [72], 


2K 


_  a2  sin  ft  sin  y  _  b2  sin  y  sin  a 


sin  a 

c2  sin  a  sin  ft 

siny 


sin  ft 


Fig.  51. 


We  have  from  geometry 

K  =  rs. 
Combining  [74]  and  [66~], 


K  =  -\/s(s-a)(s-  b)  (s  -  c). 


[73] 


[74] 
[75] 


EXERCISES 

Find  the  areas  of  the  following  triangles : 

1.  a  » 13,  b  =  10,  c  =  17. 

2.  a  =  143,  ft  =  100,  y  =  74°16'. 

3.  b  =  200,  a  =  47°  24',  y  =  63°  25'. 

4.  The  sides  of  a  triangle  are  175,  120,  215;    find  its 
area  and  the  radii  of  its  inscribed  and  escribed  circles. 

abc 


5.   Prove  K  = 


±R 


^A2       Jw^  A?  £C~   CL 


UxCuZ^. 


•2. 


*—*£• 


*       frVvc 


wQ^«        _ 


0  PLANE   TRIGONOMETRY 

Verify  the  following  identities  : 

6.  cos  i  a  cos  i  /3  cos  £  y  =  -^ .     (Use  [62].) 

7.  cot*acot*£cot£y  =  ~    (Use  [63].) 

2 

.    8.    cot£a  +  cot£0  +  cot£y  ==  -• 


CHAPTER  VII 
THE   SOLUTION   OF   THE   TRIANGLE 

60.   We  have  learned  in  geometry  that  a  triangle  can  be 
constructed  when  we  are  given  three  of  its  parts,  of  which 
one,  at  least,  is  a  side.     The  formulas  of  the  preceding 
chapter  enable  us  to  compute  the  values  of  the  unknown 
parts  when  we  know  the  measures  of  the  given  parts. 
The  three  given  parts  may  be  : 
I.    One  side  and  two  angles. 
II.    Two  sides  and  the  included  angle. 

III.  Two  sides  and  the  angle  opposite  one  of  them. 

IV.  Three  sides. 

Formulas  [57],  [58],  and  [60]  are  sufficient  to  solve  all 
four  cases.  In  the  computations  in  Chapter  II  we  used 
natural  functions ;  here  we  propose  to  use  logarithms,  and 
formula  [60]  is  not  adapted  to  logarithmic  calculation. 
In  its  place  we  shall  use  formulas  [65^  and  [66^,  which 
are  derived  from  it. 

The  necessary  formulas  are  : 

=  M,  [57] 

tan  i  (a  +  p),  [58] 

[65] 
where  r^('-»H«-»)Eg.  [66] 


a             b 

c 

sin  a      sin  f3 

sin  y 

tan  %  (a  —  (2)  = 

a  —  b 
a  +  b 

tan  ia  = 

r 

PLANE   TRIGONOMETRY 


ho 
o 


CD 


■-a 

CD 


r-  *» 


02. 


II 


«    e 


^1^    9 


00. 

d 


5      =8 


>"i      CO 

d      d 


CO. 
+ 

T 

o 

O 


Q5. 

+ 


+ 

00. 
+ 


+ 

CO. 

o  ««  »  a 


<m    ,_,    n  ~r     I 


1 


<2  J> 


c6 
© 

CD 


I 


oo. 

+ 


CO. 

+ 


e  I  8 


00. 

I 
g 


oo. 

I 


00. 

+ 


00. 


M  3 
.9  Ls 

*co  I  "53 

8 


d 


d 


8 


O 


xc2  or 

"      »       M 


o£ 


+ 

S        I  « 

i  e  Ls 

o  I  cc 

§      II 

II        ^ 


00. 

d 


d 


*? 

? 

+ 

1 

<*- 

CO 

00 

d 

OG 

H" 

H» 

00 

a 

d 

cr; 

a 

+2 

+j 

DO 

+   I 


o 


THE    SOLUTION   OF   THE    TRIANGLE 


91 


GO 

O 

o 

CQ 

O 

t— i 

« 

H 

M 

P3 

O 
O 

3 


1— 1 
H 

+           .               U 

-*          ^  111 

*  5s  oa.  "3  <**•  M 
^  +    1     o  •§    n 

.    +  ^  ^    1     op 

1  X  g>^  *•  ,  * 

. S   i    i  I'ssf  g 

*°  ^    e        ^  to  ^  & 
.   II  >S       +  o    ©    o 

+  71       ^  e   «   £ 

S  _       II   g5  S   « 

1          ^    II     o°  g> 
g                   be 

■M                                                       r— 

be 

> 

N 

roTl 

bo 

o 

I    e  S"  ^ 

~JL  i   i   i 

'     be  be  be            o 

^    no     O     O     O                   00 

to  be    1      i     i               .. 

w   ~* '   ^     S-     **     s*                  ^ 

^  +  +   be  be  be 

^  f    U  7\  li      <3a- 

g5       a   c  a 

tZll                  +J       +3       +J 

H«           bo    bo   bo 

£      r2      r2 

be 
c 

i— i 

+ 

s 

tf »  +;+^ 

^-  o     e       ^  ^    ,   h*, 

II    II          II    II  ^toto 

w          be  he    |      | 
bo         o    o     '      ' 

o      ^  ^  tr 
+ 

be 
o 

M 

1— 1 

2      ^ 

+       + 

*w       ^    .2   "   «   be^ 

bO           bO    a-  ^     gJ    W)    O    T 

^    1          1   -3    £  +  +    "   S) 

^  S        ^fe^bc^    1     1 

H        I'  %  +  ||  U  bS 

^^     z 

ii         To 

09 

O 

Data 
Solution 

Check 

CO 

O 

Data 
Solution 

Check 

92  PLANE   TRIGONOMETRY 

61.  Logarithmic  Functions.  Tables  of  logarithmic  func- 
tions are  arranged  like  tables  of  natural,  functions.  They 
consist  of  the  logarithms  of  the  natural  functions.  When, 
however,  the  characteristic  is  negative,  10  is  added.  For 
this  reason  the  characteristics  of  all  sines  and  cosines, 
of  tangents  of  angles  less  than  45°,  and  of  cotangents  of 
angles  greater  than  45°,  are  10  too  large.  This  fact  must 
be  kept  in  mind  when  computing.  A  little  experience  will 
correct  any  liability  to  error  from  this  source.  Sines  and 
tangents  of  very  small  angles,  cosines  and  cotangents  of 
angles  near  90°,  cannot  be  accurately  obtained  by  interpo- 
lation. Supplementary  tables  are  generally  furnished  for 
this  purpose. 

62.  The  actual  work  of  computation  in  each  case  will 
now  be  illustrated  by  the  solution  of  specific  problems. 
The  first  step  in  the  solution  of  every  problem  is  the 
careful  construction  of  the  figure  and  the  graphic  solution 
by  measurement.  The  results  so  obtained  serve  as  a  rough 
estimate  of  what  is  to  be  more  accurately  determined  by 
computation. 

In  the  following  illustrative  problems  the  work  is  ar- 
ranged in  convenient  form,  and  this  form  should  be  fol- 
lowed by  the  student. 

,     Case  I.     Two  Angles  and  a  Side. 

Given  a  =  571,  a  =  57°  21  '.3,  (3  =  43°  16 '.8,  find  the 
other  parts. 

f  a  =  571. 
Data  J  a  =57°  21  '.3. 

[£=43°16'.8.  Check 

y=79°21'.9.  e  +  b  =  1131.41. 

log  a  =  2.75664.  c-b  =  201.59. 

log  sin  a  =  9.92532  -  10.  i  (y  +  P)  =  61°  19'.35. 


THE    SOLUTION   OF   THE   TRIANGLE 


93 


log  M  =2.83132. 
log  sin  p  =  9.83605  -  10. 
log  sin  y  -  9.99248  -  10. 
log  b  =  2.66737. 
log  c  =  2.82380. 
b  =  464.91. 
c  =  666.5. 


±(y-0)=18o2'.55. 
log  (c  +  $)=  3.05362. 
log  (c  -  b)  =  2.30447. 

log  quotient  =  .74915. 
log  tan  £  (y  +  p)  =  .26204. 
log  tan  |  (y  -  £)  =  9.51288  -  10. 

log  quotient  =  .74916. 


1.  a  =137.43, 

2.  a  =  437.18, 

3.  5  =  943.49, 

5.  0  =  637.23, 

6.  a  =  63.72, 

7.  6  =  6.372, 

8.  b  =  .0641, 

9.  0  =  .0037, 
10.  a  =  4.003, 


EXERCISES 

a  =43°  21 '.3, 
p  =  83°  25'. 7, 
a  =  12°  17'.6, 
P  =  102°  35'.3, 
a  =  46°  46', 
a  =  l°20', 
a  =  88°  14'.5, 
a  =  36°  17'.1, 
0  =  36°  17', 
a  =  36°  17', 


£=65°23'.5. 
y  =  73°  32'.8. 
y  =  121°  07'.2. 
y  =  80°  12  M. 
p=56°  56'. 
p=75°  40'. 
y  =  88°  14 '.2. 
y  =  53°  43'.6. 
y  =  72°  34'. 
P=10S°  5V. 


63.    Case  II.     Two  Sides  and  the  Included  Angle. 
Given  a  =  1371,  6  =  1746,  y  =  46°  30',    find   the  other 
parts. 


r  a  =  1371. 
Data -I    6  =  1746. 
(.  y  =  46°  30'. 
ft  +  a  =  3117. 
b  -  a  =  375. 
£(£  +  a)  =66°  45'. 
log  (b  -  a)  =  2.57403. 


Check 
log  a  =  3.13704 
log  sin  a  =  9.89116  —  10 

3.24588 


94 


PLANE   TRIGONOMETRY 


log  tan£(0  +  a)  = 

■  .36690. 

log  b  =  3.24204 

colog  (b  +  a)  = 

=  6.50626.-  )• 

log 

sin  p  =  9.99616  - 

10 

log  tan  -|-  (/J  —  a)  = 

:  9.44719  -  10. 

3.24588 

£(/?-«)  = 

:  15°  38'.6. 

a  = 

:  51°  6'.4. 

log  c  =  3.10644 

P  = 

=  82°  23'.6. 

log 

sin  y  =  9.86056  - 

10 

loga  = 

=  3.13704. 

3.24588 

colog  sin  a  = 

:  0.10884. 

log  sin  y  = 

:  9.86056. 

logc  = 

3.10644. 

c  = 

1276.7. 

EXERCISES 

1.    a  =  127, 

b  =  145, 

y  =  24°  37'.2. 

2.    a  =127, 

b  =  145, 

y  =  84°  13'.6. 

3.    a  =  127, 

b  =  145, 

y  =  173°  28'.5. 

4.  5  =  231, 

5.  a  =231, 

c  =  31, 

a  =  74°  15'.2. 

b  =221, 

y  =  100°  14'.5. 

6.    c  =  347, 

a  =  34, 

/3=10°46'.3. 

7.    6  =  12.473, 

c  =  34.257, 

a  =  146°  24 '.1. 

8.    a  =  100, 

b  =  200, 

y  =  100°. 

9.    a  =100, 

b  =  200, 

y  =  10°. 

10.  The  line  AB  is  divided  at  D  into  two  segments, 
AD  =  200,  DB  =  100  ;  from  C  each  of  these  segments 
subtends  an  angle  of  35°.     Find  the  angles  CAB  and  CBA. 

sy  64.  Case  III.  Two  Sides  and  an  Angle  Opposite  One  of 
Them.  This  case  sometimes  admits  of  two  solutions.  Let 
the  given  parts  be  a,  b,  a.  Construct  the  angle  a.  On 
one  side  lay  off  A  C  =  b  ;  from  C  as  center  with  radius  a, 
describe  an  arc,  cutting  the  other  side  AM  at  J5X  and  B2. 


THE   SOLUTION   OF   THE   TRIANGLE 


95 


The  triaogles  ABXC  and  AB2C  both  satisfy  the  conditions, 
and  both  are  therefore  solutions.     Study  of  the  diagram 
will  show  that  we  shall  have 
two  solutions  when,  and  only 
when, 

a<90°,  b>a>:p. 

In  any  particular  case  the  FlG  52 

graphic   solution    will  deter- 
mine whether  there  is  one  or  two  solutions. 

The  angles  at  B1  and  B2  are  obviously  supplementary. 
In  the  computation  we  find  sin  /3.  Now  we  learned  in  §  26 
that  there  were  two  angles  less  than  180°  with  the  same 
sine,  the  one  the  supplement  of  the  other  ;  when  we  find 
/3  from  sin  /?,  we  must  therefore  take  not  only  the  value 
given  in  the  table  but  also  the  supplement  of  this  value. 
If  there  is  but  one  solution,  later  steps  in  the  computation 
will  compel  the  rejection  of  the  second  of  these  values. 

Given,  1.   a  =  44.243,     b  =  30.347,     a  =  34°  23'.2. 
2.    a  =  44.243,     5  =  60.347,     a  =  34°  23'.2. 


1. 

2. 

Data  ^  b 

44.243, 

44.243. 

30.347, 

60.347. 

[a 

34°  23'.2, 

34°  23'.2. 

log  a 

1.64585, 

1.64585. 

log  sin  a 

9.75188  - 

10, 

9.75188  -  10. 

log  M 

1.89397, 

1.89397.  * 

log  b 

1.48212, 

1.78066. 

log  sin  ft 

9.58815  - 

10, 

9.88669  -  10. 

P 

22°  47'. 5, 

50°  23U> 

2'. 


129°  36'.9. 


96 


PLANE   TRIGONOMETRY 


y  122°  49'.3,         95°  13'. 7,  15°59'.9. 

log  sin  r  9.92447  -  10,  9.99819  -  10,  9.44030  -  10. 

log  c  1.81844,  1.89216,  1.33427. 

c  65.833,  78.012,  21.591. 


Check 

c  +  b 

96.180, 

138.36, 

81.938. 

c-b 

35.486, 

17.665, 

38.756. 

t(f  +  P) 

72°  48'. 4, 

72°  48'. 4, 

72°  48'.4 

i(y-» 

50°  00'.9, 

22°  25'. 3, 

56°  48 '.5. 

log(c  +  ft)* 

1.98308, 

2.14101, 

1.91349. 

log  (p  -  b) 

1.55006, 

1.24712, 

1.58833. 

log  quotient 

.43302, 

.89389, 

.32516. 

log  tan          c 

.50945, 

.50945, 

.50945. 

log  tan  y  2  P 

.07641, 

9.61555  - 

-10,    .18431. 

log  quotient 

.43304, 

.89390, 

.32514. 

1.  a  =  145, 

2.  a  =  2.37, 

3.  ft  =  147.3, 

4.  a  =  32.14, 

5.  ft  =  13.47, 

6.  ft  =  .149, 

7.  a  =  1.243; 

8.  a  =  432.1, 

9.  o  =  .0027, 
10.  a  =  124, 


EXERCISES 
ft    =160, 

c   =3.14, 
a  =  124.2, 

ft'  =  270, 
e  =18.75, 
c  =.137, 
ft  =2.345, 
ft  =321.4, 
a  =  .0031, 
ft   =83, 


a  =  47°  38'. 
y  =  65°  23'. 
P  =  142°  17'. 
£  =  75°  48'.3. 
P  =  110°  43'. 
y  =  38°  47'. 
a  =  10°  57 '.5. 
0  =  28°  47'. 
a  =  84°  21'.6. 
p  =  68°  43'. 


THE    SOLUTION   OF   THE   TRIANGLE  97 

11.  I  =  241,  m  =  214,                fi  =  43°  27'. 

12.  p  =13.17,  ?  =17.13,             Q  =  71°31'. 

13.  a  =  187.5,  b   =  201.1,             a  =  67°  47'.4. 

14.  a  =  5872,  ft  =  7857,      £  =  78°  5'. 

15.  a  -  1,  b   =  2,        a  =  23°  32'. 

16.  a  =  .0003,  ft  =  .0004,     a  =  50°  5'. 

17.  a  =  3000,  ft  =  4000,      a  =    5°  50'. 

18.  a  =  1241,  ft  =  2114,      a  =  63°  36!. 

19.  ^  =  1899,  ft  =2004,      a=73°l'. 

20.  ft  =  173,  a  =  74°  12';    find  the  limits  of  a  for  two 
solutions. 

21.  <x  =  127,    ft  =  143 ;    find    the    limits    of    a   for   two 
solutions. 

65.    Case  IV.     Three  Sides.     Given  a  =  1573,  ft  =  2044, 
c  =  2736. 

r  a  =  1573,  colog  5  =  6.49805. 

Data  \  ft  =  2044,  log  (5  -  a)  =  3.20507. 

L  c  =  2736,  log  (s  -  ft)  =  3.05404. 

25  =  6353,  log  (s  -  c)  =  2.64395. 

*  =  3176.5,  log  r2  =  5.40111.     X,^ 

5  _  a  _  1603.5,  log  r  =  2.70056. 

s  -  ft  =  1132.5,  log  tan  £  a  =  9.49549  -  10. 

5  _  c  =  440.5,  log  tan  £  £  =  9.64652  -  10. 

ia  =  ir22'.7,  ^g  tan  iy  =  . 05661. 

i  fi  =  23°  53'.9, 

1-  y  =  48°  43'. 4,  Check 

a  =  34°  45'.4,  a  +  £  +  y  =  180°  OO'.O. 

0  =  47°  47'.8. 

y  =  97°  26'.8. 


98  PLANE   TRIGONOMETRY 


EXERCISES 

-  1. 

a  =  51, 

6  =  65, 

c  =  60. 

2. 

a  s=  51, 

b  =  65, 

c  =  20. 

3. 

a  =  431, 

b  =  440, 

c  =  25. 

<~  4- 

a  =  78.43, 

b  =  101.67, 

c  =  29.82 

5. 

a  =  111.1, 

b  =  120, 

c  =  130. 

A 

a  =  .003, 

6  =  .007, 

c  =  .011. 

8. 

a  =  .431, 

b  =  .34, 

c  =  .7. 

«  ==  6, 

b  =  6, 

c  =  2. 

9. 

a  =  6, 

b  =  6, 

c  =  ll. 

10. 

a  =  12, 

b  =  U, 

c-16. 

11. 

a  =  4, 

b  =  6, 

c  =  9. 

12. 

a  =  4, 

b  =  6, 

(5  =±=8. 

13. 

a  =  4, 

b  =  6, 

0  =  11. 

EXERCISES 

1.  One  side  of  a  triangular  lot  is  1427  ft. ;  the  adjacent 
angles  are  48°  15'  and  75°  35';  find  the  perimeter  and  the 
area. 

2.  Prove  that  the  area  of  a  quadrilateral  is  one-half  the 
product  of  its  diagonals  into  the  sine  of  the  angle  between 
them. 

3.  The  diagonals  of  a  parallelogram  are  17  ft.  and  30  ft., 
and  the  angle  between  them  is  64°  27' ;  find  the  sides  of 
the  parallelogram  and  its  area. 

4.  A  balloon  is  directly  over  a  straight  road.  From  two 
points  3  mi.  apart  and  on  opposite  sides  its  elevation 
was  found  to  be  30°  28'  and  47°  22' ;  what  was  its  height  ? 
If  the  two  points  of  observation  had  been  on  the  same  side 
of  the  balloon,  what  would  its  height  have  been  ? 


THE   SOLUTION   OF   THE   TRIANGLE  99 

5.  What  is  the  angle  between  two  faces  of  a  regular 
tetrahedron  ?    of  a  regular  octahedron  ? 

6.  Three  circles  whose  radii  are  12,  17,  and  19  are  tan- 
gent, two  and  two  externally ;  find  the  area  of  the  surface 
enclosed  by  them. 

7.  From  two  successive  mile  posts  on  a  straight  and 
level  road  the  elevation  of  the  top  of  a  hill  in  line  with 
them  is  8°  and  10°  ;  find  the  distance  and  height  of  the  hill. 

8.  The  longer  sides  of  a  parallelogram  are  18,  the 
shorter  sides  10,  and  one  diagonal  is  12  ;  find  the  other 
diagonal  and  the  angles. 

9.  The  parallel  sides  of  a  trapezoid  are  34  and  50,  the 
non-parallel  sides  20  and  25  ;  find  the  angles  and  the 
diagonals. 

10.  Two  sides  of  a  triangle  are  20  and  30,  and  the 
median  from  their  intersection  is  16  ;  find  the  base  and 
the  angles  of  the  triangle. 

11.  A  field  is  500  ft.  square  ;  a  post  stands  350  ft.  from 
one  corner  and  400  ft.  from  an  adjacent  corner  ;  what  are 
its  distances  from  the  other  two  corners,  1°,  when  it  is  within 
the  field ;  2°,  when  it  is  outside  ?  If  the  second  corner  were 
opposite  the  first  instead  of  adjacent  to  it,  what  would  the 
distances  be? 

^12.  Wishing  to  find  the  height  of  a  mountain,  I  measure 
a  line  of  600  yds.  in  the  same  vertical  plane  with  the  top 
of  the  mountain.  The  upper  end  of  this  line  is  40  ft. 
higher  than  the  lower  end,  and  the  elevation  of  the  moun- 
tain top  at  the  former  is  6°  23',  at  the  latter  3°  23' ; 
what  is  the  height  of  the  mountain  above  the  lower  end 
of  the  base  line  ?  If  the  lower  end  of  the  base  line  were 
next  to  the  mountain,  what  would  its  height  be  ? 


/. 


100  -    PLANE   TRIGONOMETRY 

13.  A  straight  and  level  road  runs  along  a  seacoast. 
From  two  points  on  this  road,  2  mi.  apart,  the  top  of 
a  lofty  mountain  is  visible  ;  what  measurements  must  I 
make  to  find  its  height  without  leaving  the  road? 

14.  The  parallel  sides  of  a  trapezoid  are  42  and  32,  one 
oblique  side  is  20,  and  it  makes  an  angle  of  65°  with  the 
longer  parallel  side ;  find  the  other  side,  the  diagonals,  and 
the  angles  ;  find  the  same  parts  if  the  oblique  side  makes  an 
angle  of  65°  with  the  shorter  parallel  side. 

15.  A  tower  50  ft.  high  has  a  mark  20  ft.  from  the 
ground.  At  what  distance  from  its  foot  do  the  two  parts 
of  the  tower  subtend  equal  angles  ?  at  what  distance  does 
the  lower  part  subtend  twice  the  angle  that  the  upper  does  ? 

16.  The  altitude  of  a  certain  rock  is  observed  to  be  47°, 
and  after  walking  1000  ft.  towards  it,  up  a  slope  of  22°,  the 
observer  finds  its  altitude  to  be  77° ;  find  the  height  of  the 
rock  above  the  first  point  of  observation. 

17.    From  two  points  A  and 
B,    5000  ft.    apart,    two    inac- 
cessible   points  P    and   Q   are 
visible.      I  find  the  angles 
PAB  =  107°  37', 
PBA  =  34°  23', 
QAB  =  ±3°  46', 

Fig.  53. 

QBA  =81°  11'; 

what  is  the  distance  from  P  to  Q,  1°,  when  both  are  on  the 
same  side  of  AB\  2°,  when  they  are  on  opposite  sides  ? 

18.  Two  flag-poles  are  203  ft.  apart.  From  the  middle 
point  of  the  line  joining  them  the  elevation  of  the  taller  is 
double  that  of  the  shorter ;  but  on  going  43^-  ft.  nearer  the 
shorter,  their  elevations  are  equal.  What  is  the  height  of 
each  ? 


THE   SOLUTION   OF   THE   TRIANGLE  101 

19.  From  the  top  of  a  hill  the  depressions  of  the  top 
and  bottom  of  a  flagstaff  25  ft.  high,  standing  at  the  foot 
of  the  hill,  are  45°  13'  and  47°  12',  respectively.  What  is 
the  height  of  the  hill  above  the  foot  of  the  flagstaff:  ? 

20.  A  column  on  a  pedestal  20  ft.  high  subtends  an 
angle  of  30° ;  on  approaching  20  ft.  nearer,  it  again  sub- 
tends an  angle  of  30°.     What  is  the  height  of  the  column  ? 

21.  From  the  middle  point  of  the  longest  side  of  the 
triangle,  whose  sides  are  10,  14,  17,  a  circle  is  described 
with  radius  12  ;  where  will  it  cut  the  other  sides  ? 

/  22.  Two  towers  stand  near  each  other  in  a  plane.  Their 
altitudes,  each  measured  from  the  base  of  the  other,  are 
46°  6'  and  33°  45',  respectively,  and  the  distance  between 
their  summits  is  87  ft.  What  is  the  height  of  each,  and 
what  is  their  distance  apart  ? 

23.  Three  circles  with  radii  16,  7,  5  touch  each  other 
externally ;  what  is  the  area  of  the  curvilinear  triangle 
so  formed  ?  If  the  two  smaller  circles  are  within  the  larger, 
what  is  the  area  of  the  curvilinear  triangle  ? 

24.  The  sides  of  a  triangle  are  20,  30,  40  ;  find  the 
lengths  of,  1°,  the  three  altitudes ;  2°,  the  three  medians ; 
3°,  the  bisectors  of  the  three  interior  angles  ;  4°,  the  bisec- 
tors of  the  three  exterior  angles ;  5°,  the  radii  of  the  cir- 
cumscribed circle,  the  inscribed  circle,  the  escribed  circles. 

25.  Near  the  foot  of  a  flagstaff,  150  ft.  high,  are  two 
posts,  A  70  ft.  north,  B  100  ft.  east.  What  is  the  shortest 
distance  from  T,  the  top  of  the  staff,  to  the  line  A  B? 
what  angle  does  this  line  make  with  the  ground? 

26.  Three  sides  of  a  convex  quadrilateral  inscribed  in  a 
circle  30  ft.  in  diameter  are  I  =  14  ft.,  m  =  18  ft.,  n  =  12  ft. ; 
find  the  fourth  side  and  the  angles  when,  1°,  I  is  the  middle 
one  of  the  three  given  sides  ;  2°,  when  m  is  the  middle  one  ; 
3°,  when  n  is  the  middle  one. 


102 


PLANE   TRIGONOMETRY 


27.  Standing  on  a  headland  250  ft.  high,  I  observe  a 
ship.  At  first  it  bears  N.N.W.,  and  its  angle  of  depression 
is  16°  8',  ten  minutes  later  it  bears  E.  by  S.  and  its  depres- 
sion is  32°  18' ;  find  what  direc- 
tion the  ship  is  sailing,  its  speed, 
and  how  near  its  course  lies  to  the 
headland. 

28.  A,  B,  and  C  are  three  buoys; 
AB  =  320  yds.,  BC  =  435  yds.,  CA 
=  600  yds.  A  ship  S  finds  that 
AB  subtends  an  angle  of  8°  and  BC 
an  angle  of  26°.  How  far  is  the 
ship  from  each  of  the  buoys  ? 
Suggestion.  Draw  a  circle  through  A,  C,  and  S,  cutting 
SB  produced  in  D.     Draw  AD  and   CD. 


Fig.  54. 


14  DAY  USE 

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OCT  1  1936 


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